Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

## AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Questions

Question 1.

Distinguish between heat and temperature. [T.S. Mar. 15]

Answer:

Heat

- It is a form of energy.
- It is the cause on the body.
- It is determined with calorimeter.
- It’s unit is Joule and Calories.

Temperature

- It is a degree of hotness or coldness of a body.
- It is an effect on a body.
- It is measured with thermometer.
- It’s unit is degree’Celsius or Kelvin or degree Fahrenheit.

Question 2.

What are the lower and upper fixing points in Celsius and Fahrenheit scales ? [T.S. Mar. 16]

Answer:

In Celsius scale, lower fixed point is ice point or 0°C and upper fixed point is steam point or 100°C. In Fahrenheit scale, lower fixed point is 32°F and upper fixed point is 212°F.

Question 3.

Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale ?

Answer:

Yes. The coefficients of expansion depends on scale of temperature because

\(\frac{\alpha}{{ }^{\circ} \mathrm{C}}=\frac{9}{5} \alpha /{ }^{\circ} \mathrm{F}\)

Question 4.

Can a substance contract on heating ? Give an example.

Answer:

Yes. Rubber, type metal, cast iron contract on heating.

Question 5.

Why gaps are left between rails on a railway track ? [A.P. Mar. 17, 16]

Answer:

The length of the rails increases in summer due to high temperature. Therefore a gap is left to allow this expansion.

Question 6.

Why do liquids have no linear and areal expansions ?

Answer:

Liquids occupy the same shape of vessel. They do not have individual length and area. Hence, liquids have no linear and areal expansions.

Question 7.

What is latent heat of fusion ?

Answer:

The amount of heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure is called the latent heat of fusion (L_{f}).

Question 8.

What is latent heat of vapourisation ? [Mar. 13]

Answer:

The amount of heat per unit mass required to change a substance from the liquid to the vapour state at the same temperature and pressure is called the latent heat of vaporisation (L_{v}).

Question 9.

What is specific gas constant ? Is it same for all gases ?

Answer:

Universal gas constant per molecular mass is called specific gas constant.

i.e., r = \(\frac{\mathrm{R}}{\mathrm{m}}\). No, it is different for different gases.

Question 10.

What are the units and dimensions of specific gas constant ?

Answer:

The units of specific gas constant is J/Kg/K. Dimensional formula is (m^{0}.L^{2}t^{-2}K^{1}).

Question 11.

Why utensils are coated black ? Why the bottom of the utensils are made of copper ?

Answer:

- Utensils are coated black, because it is a good absorber of heat.
- Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

Question 12.

State Weins displacement law. [A.P. Mar. 17]

Answer:

The wavelength (λ_{m}) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature

i.e., λ_{m} ∝ \(\frac{1}{T}\)

Question 13.

Ventilators are provided in rooms just below the roof. Why ? [Mar. 14]

Answer:

Ventilators are provided in rooms just below the roof, because the hot air escapes out and fresh air enters into the rooms due to convection.

Question 14.

Does a body radiate heat at 0 K ? Does it radiate heat at 0°C ?

Answer:

- No. A body does not radiate heat at 0k.
- Yes. A body radiate heat at 0°C.

Question 15.

State the different modes of transmission of heat. Which of these modes require medium ?

Answer:

The different modes of transmission are : 1) conduction 2) convection and 3) radiation.

Among three modes conduction and convection require medium.

Question 16.

Define coefficient of thermal conductivity and temperature gradient.

Answer:

Coefficient of thermal conductivity : It is defined as the quantity of heat energy flowing per second between the opposite faces of cube of unit side, which are maintained at unit temperature difference.

Temperature gradient: The change of temperature per unit length of conductor is called temperature gradient.

Question 17.

What is thermal resistance of a conductor ? On what factors does it depend ?

Answer:

The resistance offered by the conductor for the flow of heat is called thermal resistance (R) of the conductor. R = \(\frac{\mathrm{d}}{\mathrm{kA}}\). It depends on

- The nature of the material, i.e., thermal conductivity (K).
- d; length of conductor along which heat flows and
- A; area of cross section of the conductor.

Question 18.

State the units and dimensions of coefficient of convection.

Answer:

The unit of coefficient of convection is, wm^{-2}k^{-1}.

Dimensional formula (m.L^{0}T^{3}θ^{-1}).

Question 19.

Define emissive power and emissivity ?

Answer:

Emissive power : The energy flux emitted by unit surface area of a radiating body is known as emissive power (e_{λ}).

Missivity (e): It is defined as the ratio of the emissive power of the body to that of black body at the same temperature.

Question 20.

What is greenhouse effect ? Explain global warming. [A.P. Mar. 15, 13]

Answer:

Greenhouse effect: When the earth receives sun light, it gets heated up and emits infrared radiation into air. CO_{2}, CH_{4}, N_{2}O, O_{3}, Chlorofluoro carbon (green house gases) present in the air absorbs the heat content of infrared radiation and keeps the earth warm. This is called green house effect.

Global warming : As CO_{2} content increases, more heat is retained in the atmosphere and the temperatures all over the world increases. This is called global warming.

Effects : a) Polar ice caps melt and fed more water in rivers and seas results in severe floods, b) In some areas, water resources dry up leading to drought conditions.

Question 21.

Define absorptive power of a body. What is the absorptive power of a perfect black body ?

Answer:

Absorptive power : At a given temperature and wavelength, the ratio of the amount of radiant energy absorbed to the amount of radiant energy incident in a wavelength range is called the absorptive power at that temperature and wavelength. Absorptive power of a perfect black body is 1.

Question 22.

State Newton’s law of cooling. [A.P. Mar. 18, 16; T.S. Mar. 18]

Answer:

The rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings provided the temperature difference is small. i.e., – \(\frac{\mathrm{dQ}}{\mathrm{dt}} \alpha\left(\mathrm{T}_B-\mathrm{T}_{\mathrm{S}}\right)\)

Question 23.

State the conditions under which Newton’s law of cooling is applicable.

Answer:

- Loss of heat is negligible by conduction and only when it is due to convection.
- Loss of heat occurs in a streamlined flow of air i.e., forced convection.
- Temperature of the body is uniformly distributed over it.

Question 24.

The roof of buildings are often painted white during summer. Why ? [T.S. Mar. 17, 15]

Answer:

The roof of buildings are painted white during summer, because it reflects more heat radiations and absorbs less heat radiations. Hence we feel comfort inside the house.

Short Answer Questions

Question 1.

Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.

Answer:

Centigrade (Celsius) scale of temperature: In the Celsius scale of temperature, the lower fixed point is called the ice point and is assigned the value 0°C. The upper fixed point is called the steam point and is assigned the value 100°C. The interval between these two points (i.e., 100°C – 0 = 100°C) is subdivided into 100 equal parts each one corresponding to 1°C.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Relation between Celsius and Fahreinheit scales of temperature:

Difference of 100 Celsius degrees = Difference of 180 Fahrenheit degress

When the temperature of a body is measured on both the Celsius and Fahrenheit scales, let the readings be t_{C} and t_{F} respectively. Then

\(\frac{t_c-0}{t^{100}}=\frac{t_F-32}{180} \Rightarrow \frac{t_C}{5}=\frac{t_F-32}{9}\)

C = \(\frac{5}{9}\) (F – 32)

Question 2.

Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating ?

Answer:

Bimetallic strip:

Two similar strips of copper and steel are revitted side by side as a compound strip called bimetallic strip. It is straight at ordinary temperatures. When the bimetallic strip is heated, copper expands more than steel. So, it bends with copper on convex side. When the bimetallic strip is cooled below room temperature, copper contracts more than steel and it bends with copper on concave side. Bimetallic strip is used in refrigerators, automatic iron, fire alarm etc.

Question 3.

Pendulum clocks generally go fast in winter and slow in summer. Why ?

Answer:

The time period of a pendulum at t_{1}°C is T_{1} = 2π \(\sqrt{\frac{L_1}{g}}\) ………….. (1)

Where L_{1} is length of pendulum at t_{1}°C.

If L_{2} is length of pendulum at t_{2}°C,

then T_{2} = 2π \(\sqrt{\frac{L_2}{g}}\) ………………. (2)

\(\frac{(2)}{(1)}\) gives \(\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{L_1(1+\alpha t)}{L_1}}\)

Where a is the linear coefficient of expansion of pendulum clock. Where t = t_{2} – t_{1}

\(\frac{T_2}{T_1}=\sqrt{(1+\alpha t)}=(1+\alpha t)^{\frac{1}{2}}\)

\(\frac{T_2}{T_1}=1+\frac{\alpha t}{2}\)

\(\frac{T_2}{T_1}-1=\frac{\alpha t}{2}\)

\(\frac{T_2-T_1}{T_1}=\frac{\alpha t}{2}\)

\(\frac{T_2-T_1}{T_1}\) = time lost by pendulum per second.

Due to expansion in length, during summer, time period increases or the clock looses time in summer. In winter due to fall in temperature, the length contracts, time period decreases, hence clock gains time.

Question 4.

In what way is the anomalous behaviour of water advantageous to acquatic animals ? [A.P. Mar. 18, Mar. 14]

Answer:

In cold countries, as atmospheric temperature decreases, the upper layers of the lakes, rivers etc., cool, contract and sink to the bottom (fig). This goes on until the whole of the water reaches the temperature of 4°C. When the top layers cool further temperature falls below 4°C, it expands and becomes lighter. It does not sink downwards and remains at the top. With further cooling the top layer gradually form ice at the top (fig). Ice and water are bad conductors of heat. So the lower layers are protected against freezing by the layers of ice and cold water at 1°C, 2°C and 3°C. This results in water remaining at the bottom at 4°C. So that aquatic animals survive in those layers of water.

Question 5.

Explain conduction, convection and radiation with examples. [TS – Mar. ’18, ’16, ’15; AP – Mar. ’15]

Answer:

The heat is transmitted in three types. They are 1) Conduction 2) Convection 3) Radiation.

- Conduction : The process of transmission of heat from one place to other without actual movement of the particles of the medium is called conduction.

Ex : When long iron rod is heated at one end, heat transmits to the other end. - Convection : The process of transmission of heat from one place to another by the actual

movement of the particles is called convection.

Ex. : If water in a beaker is heated, the particles of water at the bottom receive the heat first. These particles expand, become lighter and rise up. At the same time colder and denser particles reach the bottom. They get in their turn heated and move up. This process is known as convection. - Radiation : The process of transmission of heat from one place to another without any intervening medium is called radiation.

Ex. : Earth receives heat radiations from the sun.

Long Answer Questions

Question 1.

State Boyle’s law and charle’s law. Hence, derive ideal gas equation which of two laws is better for the purpose of Thermometry and why ?

Answer:

P ∝ \(\frac{1}{\mathrm{V}}\)

⇒ PV = constants ⇒ P_{1}V_{1} – P_{2}V_{2}

Charles law at constant volume : At constant volume the pressure of given mass of a gas is directly proportional to absolute temperature of gas.

P ∝ T

⇒ \(\frac{P}{T}\) = constant ⇒ \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)

Charles law at constant pressure : At constant pressure the volume of given mass of gas is directly proportional to absolute temperature.

V ∝ T

⇒ \(\frac{V}{T}\) = constant ⇒ \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

Ideal gas equation : A gas which obeys all the gas laws at all temperature and all pressures is called an ideal gas.

Consider are mole of a gas initially at a pressure P_{1} volume V_{1} and temperature T_{1} and Let P_{2}, V_{2} and T_{2} be the final pressure, volume and absolute temperature T_{2}.

From Boyle’s law, P_{1}V_{1} = P_{2}V_{2}

⇒ V = \(\frac{P_1 V_1}{P_2}\) ………….. (1)

Now the gas is heated at constant pressure P_{2}, then its volume changes from V to V_{2} and temperature changes from T_{1} and T_{2}.

From charles law, \(\frac{V}{T_1}=\frac{V_2}{T_2} \Rightarrow V=\frac{V_2 T_1}{T_2}\) ………….. (2)

From (1) & (2) \(\frac{P_1 V_1}{P_2}=\frac{V_2 T_1}{T_2} \Rightarrow \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

⇒ \(\frac{P V}{T}\) = constant ⇒ \(\frac{P V}{T}\) = R

Where ‘R’ is coniversal gas constant.

⇒ PV = RT

From ‘n’ gram moles PV = nRT. This law is called ideal gas equation.

Out of Boyle’s law, charles law is better for the purpose of thermometry because with increasing temperature, pressure and volume of gas also increase.

Question 2.

Explain thermal conductivity and coefficient of thermal conductivity. A copper bar of thermal conductivity 401 W/(mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross – sectional area is 1.0 × 10^{-6} m^{-2}. What is the rate of heat conduction, along the bar ?

Answer:

Thermal conductivity : Thermal conductivity of a solid body is ability to conduct heat in the body. Coefficient of thermal conductivity: The coefficient of thermal conductivity of a material may be defined as the quantity of heat energy flowing per second between the opposite faces of the cube of unit side, which are maintained at unit temperature difference.

Explanation of conduction : In steady state, conduction between the opposite faces which are at temperatures θ_{1}°C and θ_{2}°C separated by distance d is

- Directly proportional to area of cross section of the rod.

Directly proportional to temperature difference (θ_{2}– θ_{1}) between the opposite faces. - Time of flow of heat, t
- Inversely proportional to the separation of faces ‘d’.

∴ Q ∝ \(\frac{\mathrm{A}\left(\theta_2-\theta_1\right) t}{\mathrm{~d}}\)

or Q = \(\frac{K A\left(\theta_2-\theta_1\right) t}{d}\)

K is called coefficient of thermal conductivity of the material.

When A = 1; (θ_{2}– θ_{1}) = 1; t = 1; d = 1

∴ K = Q

Problem:

K_{c} = 401 W/mK, θ_{2} = 104°C, θ_{2} = 24°C, d = 0.10 m, A = 1.6 × 10^{-6} m^{-2}

Rate of heat conduction = \(\frac{Q}{t}=K_c A \frac{\left(\theta_2-\theta_1\right)}{d}\)

= 401 × 1 × 10^{-6} × (104-24) = 0.32 W

Question 3.

State and explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable. A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings.

Answer:

Expression for Newton’s law of cooling : Consider a hot body at temperature T. Let T_{0} be the temperature of its surroundings. According to Newton’s law of cooling,

Rate of loss of heat ∝ Temperature difference between the body and its surroundings.

\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) ∝ (T – T_{0})

\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) = k (T – T_{0}) ⇒ (1) where k is proportionality constant. Let m be the mass and C the specific heat of the body at temperature T. If the temperature of the body falls by small amount dT in time dt, then the amount of heat lost is

dQ = mc dT

∴ Rate of loss of heat is given by

\(\frac{\mathrm{dQ}}{\mathrm{dt}}\) = mc \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ………….. (2)

putting (2) in (1) ⇒

-mc \(\frac{\mathrm{dT}}{\mathrm{dt}}\) = k(T – T_{0})

\(\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{mc}}\) (T – T_{0}) = -K(T – T_{0})

where K = \(\frac{\mathrm{k}}{\mathrm{mc}}\) is another constant.

⇒ \(\frac{d T}{\left(T-T_0\right)}\) = – K dt

on integrating both sides, we get

\(\int \frac{d T}{\left(T-T_0\right)}=-K \int d t\)

log_{e} (T – T_{0}) = – Kt + c …………… (3)

T – T_{0} = e^{-Kt + c}

T = T_{0} + e^{c} e^{-Kt}

T = T_{0} + Ce^{-Kt} ……………… (4)

Where c is integration constant and C = e^{c}

equations (1), (2), (3) and (4) are the different expressions for Newton’s law of cooling.

Explanation of Graphs:

1) If we plot a graph by taking equation (4) different values of temperature difference ∆T = T – T_{0} along y – axis and the corresponding values of t along X – axis, we get a curve of the form shown in figure. It clearly shows that the rate of cooling is higher initially and then decreases as the temperature of the body falls.

Curve showing cooling of hot water with time.

2) More over, the equation (3) is of the form y = mx + c. So if we plot a graph, by taking log_{e}(T – T_{0}) along Y – axis and time t along x – axis, we must get a straight line, as shown in figure. It has a negative slope equal to – K and intercept on Y – axis equal to C.

In both of the above situations, Newton’s law of cooling stands verified.

Newton’s law of cooling is applicable :

- Loss of heat is negligible by conduction and only when it is due to convection.
- Loss of heat occurs in a stream lined flow of air i.e., forced convection.
- Temperature of the body is uniformly distributed over it.
- Temperature differences are moderate i.e., upto 30 K, however if heat body is due to forced convection the law is valid for large differences of temperature also.

∴ Surrounding temperature, θ_{0}= \(\frac{85^{\circ} \mathrm{C}}{3}\) = 28.33°C

Problems

Question 1.

What is the temperature for which the readings on Kelvin and Fahrenheit scales are same?

Answer:

The relation between Kelvin and Fahrenheit scales is given by

\(\frac{K-273.15}{100}=\frac{F-32}{180}\)

but K = F

\(\frac{F-273.15}{100}=\frac{F-32}{180}\)

F – 273.15 = \(\frac{5}{9}\) F – \(\frac{160}{9}\)

F – \(\frac{5}{9}\) F = 273.15 – \(\frac{160}{9}\)

= 273.15 – 17.77

\(\frac{4F}{9}\) = 255.38

∴ F = \(\frac{9}{4}\) (255.38)

= 574.6°F

Question 2.

Find the increase in temperature of aluminium rod if its length ¡s to be increased by 1%. (a for aluminium = 25 × 10^{-6/0} C). [A.P (Mar. ‘15)]

Answer:

Percentage increase in length = \(\frac{l_2-l_1}{l_1}\) × 100

= α(t_{2} – t_{1}) × 100 [∵ \(\frac{l_2-l_1}{l_1}\) = α(t_{2} – t_{1})]

Here, percentage increase in length = 1,

α = 25 × 10^{-6}/ °C

1 = 25 × 10^{-6}(t_{2} – t_{1}) × 100

t_{2} – t_{1} = \(\frac{1}{25 \times 10^{-4}}=\frac{10^4}{25}\) = 400°C

Question 3.

How much steam at 100°C is to be passed into water of mass 100 g at 20°C to raise its temperature by 5°C ? (Latent heat of steam is 540 cal/g and specific heat of water is 1 cal/g°C).

Answer:

In the method of mixtures,

Heat lost by steam = Heat gained by water

m_{S}L_{S} + m_{S}S(100 – 1) = m_{W}S (t – 20)

Where m_{S} is the mass of steam, L_{S} is the latent heat of steam, S is the specific heat of steam and m_{W} is the mass of water.

Here, L_{S} = 540 cal/g

S = 1 cal/g°C;

m_{W} = 100 g

t = 20 + 5 = 25°C

m_{S} × 540 + m_{S} × 1 × (100 – 25) = 100 × 1 × (25 – 20)

615 m_{S} = 500

m_{S} = \(\frac{500}{615}\)

= 0.813 g.

Question 4.

2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kgK, find the amount of heat absorbed in kJ and kcal.

(J = 4.2 kJ/kcal).

Answer:

Here, M = 2 kg

dT = 313- 293 = 20 K

C_{V} = 0.718 × 10^{3} J/Kg – K

J = 4.2 KJ/Kcal

C_{V} = \(\frac{1}{\mathrm{~m}} \frac{\mathrm{d} \theta}{\mathrm{dt}}\)

dθ = mC_{V}dT

= 0.718 × 10^{3} × 2 × 20

= 28720 J

dθ = 28.72 KJ

dθ = \(\frac{28720}{4.2 \times 10^3}\)

= 6.838 Kcal.

Question 5.

A dock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.

Answer:

Here t_{1} = 20°C, t_{2} = 30°C

Loss of time per day = 8.212 sec

Loss of time per day = \(\frac{1}{2}\) α (t_{2} – t_{1}) × 86,400

8.212 = \(\frac{1}{2}\) α (30 – 20) × 86,400

∴ α = \(\frac{16.424}{864 \times 10^3}\)

= 19 × 10^{-6}/C°

Question 6.

A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C ?

Answer:

Rate of cooling \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K(θ_{av} – θ_{0})

Case (i) :

Given, .

θ_{1} = 60°C, θ_{2} = 40°C, θ_{0} = 10°C, t = 7 min

⇒ \(\frac{60^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}}{7}=K\left[\frac{60^{\circ} \mathrm{C}+40^{\circ} \mathrm{C}}{2}-10^{\circ} \mathrm{C}\right]\)

\(\frac{20}{7}\) = K[50 – 10]

⇒ \(\frac{20}{7}\) = K × 40

∴ K = \(\frac{1}{14}\) ……………. (1)

Case (ii) :

Given,

θ_{1} = 40°C, θ_{0} = 10°C, t = 7 min, θ_{2} = ?

⇒ \(\frac{40-\theta_2}{7}=K\left[\frac{40+\theta_2}{2}-10\right]\)

\(\frac{40-\theta_2}{7}=\frac{1}{14}\left[\frac{40+\theta_2}{2}-10\right]\)

80 – 2θ_{2} = \(\frac{40+\theta_2-20}{2}\)

160 – 4θ_{2} = 20 + θ_{2}

5θ_{2} = 140

∴ θ_{2} = \(\frac{140}{5}\) = 28°C

Question 7.

If the maximum intensity of radiation for a black body is found at 2.65 pm what is the temperature of the radiating body ? (Wien’s constant = 2.9 × 10^{-3} mK)

Answer:

λ_{max} = 2.65 pm = 2.65 × 10^{-6}m

Wien’s constant (b) = 2.90 × 10^{-3} mk

λ_{max} T = b(constant)

T = \(\frac{b}{\lambda_{\max }}=\frac{2.9 \times 10^{-3}}{2.65 \times 10^{-6}}\)

= 1094 K.

Additional Problems

Question 1.

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer:

Relation between kelvin scale and Celsius scale is T_{C} = T_{K} – 273.15

Where T_{C}.T_{K} = temperature on Celsius and kelvin scales respectively

For Neon T_{C} = 24.57 – 273.15 = – 248.58°C

For CO_{2} T_{C} = 216.55 – 273.15 = -56.60°C

Relation between Kelvin and Fahrenheit scales is

\(\frac{T_F-32}{180}=\frac{T_K-273.15}{100}\)

T_{F} = \(\frac{180}{100}\) (T_{K} – 273.15) + 32

FQ or Neon T_{F}

= \(\frac{180}{100}\)(24.57 – 273.15) + 32 = – 415.44° F

FQ or CO_{2} T_{F} = \(\frac{180}{100}\) (216.55 – 273.15) + 32

= – 69.88° F

Question 2.

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_{A} and T_{B} ?

Answer:

Given triple point of water on scale A = 200

A triple point of water on scale B = 350 B

As per question 200 A = 350, B = 273.16 K

(Or)

IA = \(\frac{273.16}{200}\) K and IB = \(\frac{273.16}{350}\) K

If T_{A} and T_{B} represent the triple point of water on two scales A and B then

\(\frac{273.16}{200}\) T_{A} = \(\frac{273.16}{350}\)T_{B} (Or) \(\frac{T_A}{T_B}=\frac{200}{350}=\frac{4}{7}\)

(Or) T_{A} = \(\frac{4}{7}\) T_{B}

Question 3.

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

R = R_{0}[1 + α(T – T_{0})]

The resistance is 101.6 Ω at the triple-point of water 273.16 K and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?

Answer:

Here, R_{0} = 101. 6 Ω, T_{0} = 273.16 K

Case (i) R_{1} = 165.5 Ω, T_{1} = 600.5 K

Case (ii) R_{2} = 123.4 Ω; T_{2} = ?

Using the relation R = R_{0}(1 + α(T – T_{0}))

Case (i):

165.5 = 101.6(1 + α(600.5-273.16))

α = \(\frac{165.5-101.6}{101.6 \times(600.5-273.16)}\)

= \(\frac{63.9}{101.6 \times 327.34}\)

Case (ii): 123.4 = 101.6(1 + α(T_{2} – 273.16))

(Or)

123.4 = 101.6

(1 + \(\frac{63.9}{101.6 \times 327.34}\) (T_{2} – 273.16))

= 101.6 + \(\frac{63.9}{327.34}\) (T_{2} – 273.16)

T_{2} = \(\frac{(123.4-101.6) \times 327.34}{63.9}\) + 273.16

= 111.67 + 273.16 = 384.83 K

Question 4.

Answer the following :

a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?

b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?

c) The absolute temperature (Kelvin scale) T is related to the temperature t_{C} on the Celsius scale by

t_{c} = T – 273.15

Why do we have 273.15 in this relation and not 273.16 ?

Answer:

a) This is on account of the fact that the triple point of water has a unique value i.e., 273.16 K at a unique point, where exists unique values of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of value as they change in pressure and volume.

b) The other fixed point on the absolute scale is the absolute zero itself.

c) On Celsius 0°C corresponds to melting points of ice of normal pressure. The corresponding value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.

From the given relation the corresponding value of triple point of water on Celsius scale

= 273.16 – 273. 15 = 0.01°C.

Question 5.

Two. ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

b) What do you think is the reason behind the slight difference in answers of thermoineters A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?

Answer:

a). Let T be the melting point of sulphur, for water T_{tr} = 273.16 K

For thermometer A, T = P/P_{tr} × 273.16

= \(\frac{1.797 \times 10^5}{1.250 \times 10^5}\) × 273.16 = 392.69 K

For thermometer B, T = P/P_{tr} × 273.16

= \(\frac{0.287 \times 10^5}{0.200 \times 10^5}\) × 273.16 = 391.98 K

b) The cause of slight different answers is that the oxygen and hydrogen gases are not perfectly ideal. To reduce this discrepancy, the readings should be at lower and lower pressure as in that case, the gases approach to the ideal gas behaviour.

Question 6.

A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0°C ? Coefficient of linear expansion of steel = 1.20 × 10^{-5} K^{-1}.

Answer:

Length of steell tape at 27°C is 100 cm i.e.

L = 100 cm and T = 27°C

The length of steel tape at 45°C is L^{1} = L + ∆L

= L + αL∆T

= 100 + (1.20 × 10^{-5}) × 100 × (45° – 27)

= 100.0216 cm

Length of 1 cm mark at 27°C on this scale, at 45°C

= 100.0216/100 cm

Length of 63 cm measured’ by this tape at 45°C will be

= \(\frac{100.0216}{100}\) × 63 = 63.0136 cm

Length of the same steel rod on a day when the temperature is 27°C = 63 × 1 = 63 cm.

Question 7.

A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :

α_{steel} = 1.20 × 10^{-5}K^{-1}.

Answer:

Here T_{1} = 27°C = 27 + 273 = 300 K

Length at temperature, T_{1}K = L_{T1} = 8.70 cm

Length at temperature, T_{2}K = L_{T2} – 8.69 cm

Change in length = L_{T2} – L_{T1} = L_{T2} α(T_{2} – T_{2})

Or 8.69 – 8.70 = 8.70 × (1.20 × 10^{-5}) × (T_{2} – 300)

Or T_{2} – 300 = \(\frac{0.01}{8.70 \times 1.2 \times 10^{-5}}\) = -95.8

Or T_{2} = 300 – 95.8 = 204.2 K = -68.8°C

Question 8.

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C ? Coefficient of linear expansion of copper = 1.70 × 10^{-5}K^{-1}.

Answer:

In this problem superfical expansion of copper sheet will be involved on heating. Here, area of hole at 227°C, then area of the hole at 227°C,

S_{2} = \(\frac{\pi \mathrm{D}_2^2}{4}\)cm^{2}.

Coefficient of superficial expansion of copper is β = 2α = 2 × 1.70 × 10^{-5} = 3.4 × 10^{-5} c^{1}

Increase in area = S_{2} – S_{1} = βS_{1} ∆T (or)

S_{2} = S_{1} + βS_{1} ∆T

= S_{1}(1 + β∆T)

(Or)

\(\frac{\pi \mathrm{D}_2^2}{4}=\frac{\pi}{4}\) (4.24)^{2} [1 + 3.4 × 10^{-5} (228 – 27)]

(Or)

D_{2}^{2} = (4.24)^{2} × 1.0068

D_{2} = 4.2544 cm

Change in diameter = D_{2} – D_{1}

= 4.2544 – 4.24

= 0.0144 cm.

Question 9.

A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10^{-5}K^{-1}; Young’s modulus of brass = 0.91 × 10^{11} Pa.

Answer:

Here L = 1.8m, T_{1} = 27°C, T_{2} = -39°C, r = 1

mm = 10^{-3}m, F = ?

α = 2 × 10^{-5}C^{-1}, y = 0.91 × 10^{11} N/m^{2}

From y = \(\frac{\mathrm{FL}}{a \Delta L}\), ⇒ ∆L = \(\frac{\mathrm{FL}}{\mathrm{ay}}\)

Also ∆L = αL∆T = ∴ \(\frac{\mathrm{FL}}{\mathrm{ay}}\) = αL∆T

(Or) F = α∆Tay = α(T_{2} – T_{1}) πr^{2}y

= 2 × 10^{-5} × (-39 – 27) × (10^{-3})^{2} × 0.91 × 10^{11}

= -3.77 × 10^{-2} N

Negative sign indicates that the force is inward due to contraction of the wire.

Question 10.

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2,0 × 10^{-5}K^{-1}, steel = 1.2 × 10^{-5}K^{-1}.

Answer:

∆L_{1} = L_{1}α_{1}∆T = 50 × (2.10 × 10^{-5}) (250 – 40) = 0.2205 cm

∆L_{2} = L_{2}α_{2}∆T

= 50 × (1.2 × 10^{-5}) (250 – 40) = 0.216 cm

∴ Change in length of combined rod

= ∆L_{1} + ∆L_{2}

= 0.220 + 0.126 = 0.346 cm

Question 11.

The coefficient of volume expansion of glycerin is 49 × 10^{-5}K^{-1}. What is the fractional change in its density for a 30°C rise in temperature ?

Answer:

Here r = 49 × 10^{-5}C^{-1}, ∆T = 30°C .

As V^{1} = V + ∆V = V(1 + r∆T)

∴ V^{1} = V(1 + 49 + 10^{-5} × 30) = 1.0147 V

As P = \(\frac{m}{V}\), p^{1} = \(\frac{m}{V^1}\) = \(\frac{\mathrm{m}}{1.0147 \mathrm{~V}}\)= 09855 P

Fractional change in density = \(\frac{\rho-\rho^{\prime}}{\rho}\)

= \(\frac{\rho-0.9855 \rho}{\rho}\)

= 0.0145

Question 12.

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg^{-1}K^{-1}.

Answer:

Here p = 10 kw = 10^{5}w,

Mass m = 8.0 kg = 8 × 10^{3} g

Rise in temp; ∆T = ?

time t = 2.5 min = 2.5 × 60 sec

Sp. heat C = 0.91 Jg^{-1}°C^{-1}

Total energy = p × t = 10^{4} × 150

= 15 × 10^{5} J

As 50% of energy is lost

∴ Energy available ∆Q = \(\frac{1}{2}\) × 15 × 10^{5}

= 7.5 × 10^{5} J

As ∆Q = mc ∆T

∴ ∆T = \(\frac{\Delta Q}{\mathrm{mc}}=\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91}\) = 103°C

Question 13.

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper = 0.39 Jg^{-1}K^{-1} heat of fusion of water = 335 Jg^{-1}.

Answer:

Here, mass of copper block m = 2.5 kg = 2500 kg

Fall in temperature ∆T = 500 – 0 = 500°C

Specific heat of copper C = 0.39 Jg^{-1}°C^{-1}

Latent heat of fusion L = 335 Jg^{-1}

Let the mass of ice melted by m^{1}

As heat gained by ice = Heat lost by copper

m^{-1}L = Mc∆T

m^{-1} = \(\frac{\mathrm{mC} \Delta \mathrm{T}}{\mathrm{L}}\)

m^{1} = \(\frac{2500 \times 0.39 \times 500}{335}\)

= 1500 g = 1.5 kg

Question 14.

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?

Answer:

Here, mass of metal, m = 0.20 kg = 200 g

Fall in temperature of metal ∆T = 150 – 40

= 110°C

If L is specific heat of metal, then heat lost by the metal

∆Q = mc∆T = 200 × L × 110

Volume of water = 150 C.C

Mass of water m^{1} = 150 g

Water equivalent of calorimeter, w = 0.025 kg = 25 kg

Rise in temp of water and calorimeter

∆T^{1} = 40 – 27 = 13°C

Heat gained by water and calorimeter,

∆Q^{1} = (m^{1} + w)T^{1}

= (150 + 25) × 13 = 175 × 13

As ∆Q = ∆Q^{1}

∴ From (i) and (ii)

200 × C × 110 = 175 × 13

C = \(\frac{175 \times 13}{200 \times 110}\) ≈ 0.1

(Or)

If some heat is lost to the sorroundings, value of L is so obtained will be less than the actual value of L.

Question 15.

Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?

Answer:

The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat

= \(\frac{5}{2}\) R = \(\frac{5}{2}\) × 1.98 = 4.95, which agrees fairly well with all observations listed in the . table except for chlorine. A monoatomic gas molecules has only the transnational motion. A diatomic gas molecules, a part from translation motion. The vibrational as well as rotational motion is also possible. Therefore to raise the temperature of 1 mole of diatomic gas through 1°C heat is also to be supplied increase not only translation energy but also rotational and vibrational energies. Hence molar specific heat of diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. Shows that for chlorine molecule, at room temperature vibrational motion. Also occurs along with translational and rotational motions. Where as other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This the reason that chlorine has some what larger value of molar specific heat.

Question 16.

Answer the following questions based on the P-T phase diagram of carbon dioxide :

a) At what temperature and pressure can the solid, liquid and vapour phases of CO_{2} co-exist in equilibrium ?

b) What is the effect of decrease of pressure on the fusion and boiling point of CO_{2} ?

c) What are the critical temperature and pressure for CO_{2} ? What is their significance ?

d) Is CO_{2} solid, liquid or gas at (a) – 70°C under 1 atm, (b) -60°C under 10 atm, (c) 15°C under 56 atm ?

Answer:

a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the point i.e., temperature = -56.6°C and pressure = 5.11 atm.

b) With the decrease in pressure, both the fusion and boiling point of carbondioxide will decrease.

c) For carbondioxide, the critical tempera-ture is 31,1°C and critical pressure is 73.0 atm. If the temp of CO_{2} is more than 31.1°C. It cannot be liquified, however large pressure we may apply.

d) Carbondioxide will be (a) a vapour at – 70°C under 1 atm (b) a solid at 6°C under 10 atm (c) a liquid at 15°C under 56 atm.

Question 17.

Answer the following questions based on the P – T phase diagram of CO_{2} :

a) CO_{2} at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase ?

b) What happens when CO_{2} at 4 atm pressure is cooled from room tem-perature at constant pressure ?

c) Describe qualitatively the changes in a given mass of solid CO_{2} at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.

d) CO_{2} is heated to a temperature 70°C and compressed isothermally. What changes in its properties, do you expect to observe ?

Answer:

a) Since the temp – 60°C lies to the left of 56.6°C on the curve i.e. lies in the region of vapour and solid phase. So CO_{2} will condense directly into solid with out becoming liquid.

b) Since the pressure 4 atm is less than 5.11 atm. The carbondioxde will condenses directly into solid without becoming liquid.

c) When a solid CO_{2} at 10 atm. Pressure and -65°C temp is heated, It is first converted into a liquid. A further increase in temperature brings it to the vapour phase. At P = 10 atm. If a horizontal line is drawn parallel to the T-axis. Then the points of intersection of this line with the fusion and vapourisation curve will give the fu-sion and boiling point of CO_{2} at 10 atm.

d) Since 70°C is higher than the critical temperature of CO_{2}. So the CO_{2} gas can not be converted into liquid state on being compressed isothermally at 70°C. It will remain in the vapour state. Nowever the gas will depart more and more now its perfect gas behavious with the increase in pressure.

Question 18.

A child running a temperature of 101 °F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g^{-1}.

Answer:

Here fall in temp = ∆T = 101 – 98 = 3°F

= 3 × \(\frac{5}{3}\)°C = \(\frac{5}{3}\)°C

Mass of child M = 30 kg

Sp. heat of human body = Sp heat of water

C = 1000 cal.kg^{-1}C^{-1}

∴ Heat last by the body of child, ∆Q = mC∆T

= 30 × 1000 × \(\frac{5}{3}\) = 5000 calories

If m be the mass of water evapourated in 20 min then m^{1}L = ∆Q

m^{1} = \(\frac{\Delta Q}{L}=\frac{5000}{580}\) = 86.2 g

∴ Average rate of extra evapouration = \(\frac{86.2}{20}\)

= 4.31 gmin^{-1}

Question 19.

A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C and co-efficient of thermal conductivity of thermacole is 0.01 Js^{-1}m^{-1} K^{-1}. [Heat of fusion of wafer = 335 × 10^{3} J kg^{-1}]

Answer:

Here length of each side l = 30 cm = 0.3 m

Thickness of each side ∆x = 5 cm = 0.05 m

Total surface area through which heat enters into the box

A = 6 l^{2} = 6 × 0.3 × 0.3 = 0 Jum^{2}

Temp.diff ∆T = 45 – 0 = 45°C,

K = 0.01 JS^{-1}m^{-1}°C^{-1}

Time ∆T = 6 hrs = 6 × 60 × 60 S

Latent heat of fusion L = 335 × 10^{3} J/kg

Let m be the mass of ice melted in this time

∆Q = mL = KA \(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) ∆t

m = kA \(\left(\frac{\Delta T}{\Delta \mathrm{T}}\right) \frac{\Delta \mathrm{t}}{\mathrm{L}}\) = 0.01 × 0.54 × \(\frac{45}{0.05}\) × \(\frac{6 \times 60 \times 60}{335 \times 10^3}\) = 0.313 kg

Mass of ice left = 4 – 0.313 = 3.687 kg

Question 20.

A brass boiler has a base area of 0.15 m^{2} and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js^{-1}m^{-1} K^{-1}: Heat of vaporisation of water = 2256 × 10^{3} Jkg^{-1}.

Answer:

Were A = 0.15 m^{2} ∆x = 1.0 m = 10^{-2} m

\(\frac{\Delta Q}{\Delta t}=\frac{6 \times 10^3 \times 2256}{60}\) JS^{-1} = 2256 × 10^{2} JS^{-1}

K = 609 JS^{-1}m^{-1}°C^{-1}, ∆T = (t – 100)

From \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\mathrm{KA}\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\)

2256 × 10^{2} = 609 × 0.15 \(\frac{(t-100)}{10^{-2}}\)

t – 100 = \(\frac{2256}{609 \times 0.15}\) = 2470 t

t = 24.70

Question 21.

Explain why :

a) a body with large reflectivity is a poor emitter.

b) a brass tumbler feels much colder than a wooden tray on a chilly day.

c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.

d) the earth without its atmosphere would be inhospitably cold.

e) healting systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Answer:

a) This is because a body with large refelectivity is a poor absorber of heat and poor absorbers of heat are poor imilters.

b) When we touch a brass tumbler on a chill day; heat flows from our body to the tumbler quickly cas thermal conductivity or brass is very high and as a result, it appears colder. On the other hand as the wood is bad conductor, heat does not flow to the wooden tray from our body, on touching it.

c) When the red hot iron pieces is in the oven, its temperature TK is the given by the relation.

E = αT^{4} But if the red hot iron piece is in the open having the surrounding temperature T_{0}, its energy is radiated according to relation E^{1} = α(T^{4} – T^{4}_{0}). As the working principle of optical pysometer is based on the fact that the brightness of – a glowing surface of a body depends , upon its temperature. Therefore, pysometes gives too low a value for the temperature of red iron in the open.

d) The lower layer’s of Earth’s atmosphere reflect infrared radiations from earth back to the surface of the earth. Thus the heat radiation recieved by the earth from the sun during the day are kept trapped by the atm. If atmosphere of earth were not there, its surface would become too cold to live.

e) Steam at 100°C possess more heat than the same mass of water at 100°C possess 540 calories of heat more than possessed by 1 gm of ware at 100°C. That is why heating systems based on circulation of stream are more efficient than those based on circulation of hot water.

Question 22.

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

Answer:

\(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K [θ_{av} – θ_{0}]

\(\frac{80-50}{5}\) = K(65 – 20)

⇒ \(\frac{30}{5}\) = K × 45 = K = \(\frac{2}{15}\)

\(\frac{60-30}{t}=\frac{2}{15}[45-20]\)

\(\frac{30}{t}=\frac{2}{15} \times 15\)

t = 5 min

Time of cooling is given by t = \(\frac{2.3026}{k}\) log_{10}

\(\frac{T_1-T_0}{T_2-T_0}\)

As per question condition (i) T_{1} = 80°C, T_{2} = 50°C, T_{0} = 20°C, t = 5 min

5 × 60 = \(\frac{2.3026}{K} \log _{10} \frac{80-20}{50-20}\)

= \(\frac{2.3026}{k} \log _{10^2}\) ……………. (2)

As per question condition (i) T_{1} = 60°C, T_{2} = 30°C, T_{0} = 20°C, t = ?

t = \(\frac{2.3026}{K} \log _{10} \frac{60-20}{30-20}\)

= \(\frac{2.3026}{k} \log _{10^4}\) ……………. (3)

Div (3) by (2) we get

\(\frac{t}{5 \times 60}=\frac{\log _{10^4}}{\log _{10^2}}=\frac{0.6021}{0.3010}\) = 2

(Or) t = 5 × 60 × 2 = 10 × 60 s

= 10 min

Textual Examples

Question 1.

Show that the coefficient of areal expansions. (∆A/A)/∆T, of rectangular sheet of the solid is twice its linear expansivity, α_{1}.

Answer:

∆A_{3} = (∆a) (∆b)

When the temperature increases by ∆T, a increases by ∆a = α_{1}a ∆T and b increases by ∆b = α_{1}b ∆T. From fig. the increase in area

∆A = ∆A_{1} + ∆A_{2} + ∆A_{3}

∆A = a ∆b + b ∆a + (∆a) (∆b)

= a α_{1}b ∆T + b α_{1}a ∆T + (α_{1})^{2} ab(∆T)^{2}

= α_{1}ab ∆T(2 + α_{1}∆T)

= α_{1}A ∆T(2 + α_{1}∆T)

Since α_{1} = 10^{-5} K^{-1}, from Table 12.1, the product α_{1}∆T for fractional temperature is small in comparision with 2 and may be neglected.

Hence,

\(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}\right) \frac{1}{\Delta \mathrm{T}}\) ≈ 2α_{1}

Question 2.

A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diametbr of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27°C. To what temperature should the ring be heated so as to fit the rim of the wheel ?

Answer:

Given,

T_{1} = 27°C

L_{T1} = 5.231 m

L_{T2} = 5.243 m

So,

L_{T2} = L_{T1} [1 + α_{1}(T_{2} – T_{1})]

5.243 m = 5.231 m[1 + 1.20 × 10^{-5} K^{-1} (T_{2} – 27°C)]

or T_{2} = 218°C

Question 3.

A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.

Answer:

Mass of aluminium sphere (m.,) = 0.047 kg

Initial temp, of aluminium sphere = 100°C

Final temp. = 23°C

Change in temp. (∆T) = (100°C – 23°C)

= 77°C

Let specific heat capacity of aluminium be s_{Al}. The amount of heat lost by the aluminium sphere = m_{1}s_{Al} ∆T = 0.047 kg × s_{Al} × 77°C s ……………. (i)

Mass of water (m_{2}) = 0.25 kg

Mass of calorimeter (m_{3}) = 0.14 kg

Initial temp, of water and calorimeter = 20°C

Final temp, of the mixture = 23°C

Change in temp. (∆T_{2}) = 23°C – 20°C = 3°C

Specific heat capacity of water (s_{w}) from table

12.3 = 4.18 × 10^{3} J kg^{-1} K^{-1}

Specific heat capacity of copper calorimeter = 0.386 × 10^{3} J kg^{-1} K^{-1}

The amount of heat gained by water and calorimeter = m_{2}s_{w} ∆T_{2} + m_{3}s_{cu} ∆T_{2} = (m_{2}s_{w} + m_{3}s_{cu}) (∆T_{2})

= (0.25 kg × 4.18 × 10^{3} J kg^{-1} K^{-1} + 0.14 kg × 0.386 × 10^{3} J kg^{-1} K^{-1}) (23°C – 20°C) ………………… (ii)

In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.

So, from (i) and (ii)

0.047 kg × s_{Al} × 77°C = (0.25 kg × 4.18 × 10^{3} J kg^{-1} K^{-1} + 0.14 kg × 0.386 × 10^{3} J kg^{-1} K^{-1}) (3°C)

s_{Al} = 0.911 kJ kg^{-1} K^{-1}

Question 4.

When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice.

(s_{water} = 4186 J kg^{-1} K^{-1})

Answer:

Heat lost by water = ms_{w} (θ_{f} – θ_{i})_{w}

= (0.30 kg) (4186 kg^{-1} K^{-1}) (50.0°C – 6.7°C)

= 54376.14 J

Heat required to melt ice = m_{2}L_{f} = (0.15 kg) L_{f}

Heat required to raise temperature of ice water to final temperature

= m_{I}s_{w} (θ_{f} – θ_{i})_{I}

= (0.15 kg) (4186 J kg^{-1} K^{-1}) (6.7°C – 0°C)

= 4206.93 J

Heat lost = heat gained

54376.14 J = (0.15 Kg)L_{f} + 4206.93 J

L_{f} = 3.34 × 10^{5} J kg^{-1}

Question 5.

Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter-to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg^{-1} K^{-1}, specific heat capacity of water = 4186 J kg^{-1} K^{-1}, latent heat of fusion of ice = 3.35 × 10^{5} J kg^{-1} and latent heat of steam = 2.256 × 10^{6} J kg^{-1}.

Answer:

We have

Mass of the ice, m = 3 kg

Specific heat capacity of ice, s_{ice}

= 2100 J kg^{-1} K^{-1}

Specific heat capacity of water, s_{water}

= 4186 J kg^{-1} K^{-1}

Latent heat of fusion ice, L_{f ice}

= 3.35 × 10^{5} J kg^{-1}

Latent heat of steam, L_{steam}

= 2.256 × 10^{6} J kg^{-1}

Now,

Q = Heat required to convert 3 kg of ice at -12°C to steam at 100°C.

Q_{1} = Heat required to convert ice at – 12°C to ice at 0°C. .

ms_{ice} ∆T_{1} = (3 kg) (2100 J kg^{-1} K^{-1}) [0 – (-12)]°C = 75600 J

Q_{2} = Heat required to melt ice at – 0°C to water at 0°C.

mL_{f ice} = (3 kg) (3.35 × 10^{5} J kg^{-1}) = 1005000 J

Q_{3} = Heat required to convert water at 0°C to water at 100°C.

ms_{W} ∆T_{2} = (3 kg) (4186 J kg^{-1} K^{-1}) (100°C)

= 1255800 J

Q_{4} = Heat required to convert water at 100°C to steam at 100°C.

mL_{steam} = (3 kg) (2-256 × 10^{6} J kg^{-1})

= 6768000 J

So,

Q = Q_{1} + Q_{2} + Q_{3} + Q_{4}

= 75600 J + 1005000 J + 1255800 J + 6768000 J

= 9.1 × 10^{6} J

Question 6.

What is the temperature of the steel- copper junction in the steady state of the system shown in fig. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300°C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s^{-1}m^{-1}K^{-1} and of copper = 385 J s^{-1}m^{-1}K^{-1}].

Answer:

Let T be the temperature of the steel-copper junction in the steady state. Then,

\(\frac{K_1 A_1(300-T)}{L_2}=\frac{K_2 A_2(T-0)}{L_2}\)

Where 1 and 2 refer to the steel and copper rod respectively. For A_{1} = 2, A_{2}, L_{1} = 15.0 cm, L_{2} = 10.0 cm, K_{1} = 50.2 J s^{-1}m^{-1}K^{-1},

K_{2} = 385 J s^{-1}m^{-1}K^{-1}, we have

\(\frac{50.2 \times 2(300-\mathrm{T})}{15}=\frac{385 \mathrm{~T}}{10}\)

which gives T = 44.4°C

Question 7.

An iron bar (L_{1} = 0.1 m, A_{1} = 0.02 m^{2}, K_{1} = 79 W m^{-1} K^{-1}) and a brass bar (L_{2} = 0.1 m, A_{2} = 0.02 m^{2}, K_{2} = 109 W m^{-1}K^{-1}) are soldered end to end as shown in fig. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar and (iii) the heat current through the compound bar.

Answer:

Given,

L_{1} = L_{2} = L = 0.1 m, A_{1} = A_{2} = A = 0.02 m^{2}, K_{1} = 79 W m^{-1} K^{-1}, K_{2} = 109 W m^{-1}K^{-1}, T_{1} = 373 K and T_{2} = 273 K.

Under steady state condition, the heat current (H_{1}) through iron bar is equal to the heat current (H_{2}) through brass bar.

So, H = H_{1} = H_{2}

= \(\frac{K_1 A_1\left(T_1-T_0\right)}{L_1}=\frac{K_2 A_2\left(T_0-T_2\right)}{L_2}\)

For A_{1} = A_{2} = A and L_{1} = L_{2} = L this equation leads to

K_{1}(T_{1} – T_{0}) = K_{2}(T_{0} – T_{2})

Thus the junction temperature T_{0} of the two bars is T_{0} = \(\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}\) ………….. (a)

Using this equation, the heat current H through either bar is

Using these equations, the heat current H’ through the compound bar of length L_{1} + L_{2} = 2L and the equivalent thermal conductivity K, of the compound bar are given by

Question 8.

A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long will it take to cool from 71°C to 69°C ?

Answer:

The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature. Under these conditions the pan cools 8°C in 2 minutes.

Using \(\frac{\mathrm{dT}_2}{\mathrm{~T}_2-\mathrm{T}_1}=-\frac{\mathrm{k}}{\mathrm{ms}}\) dt = – K dt we have

The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.

The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.

\(\frac{2^{\circ} \mathrm{C}}{\text { Time }}\) = K(50°C)

When we divide above two equations, we have

\(\frac{8^{\circ} \mathrm{C} / 2 \min }{2^{\circ} \mathrm{C} / \text { time }}=\frac{K\left(70^{\circ} \mathrm{C}\right)}{\mathrm{K}\left(50^{\circ} \mathrm{C}\right)}\)

Time = 0.7 min = 42 s