WBCHSE Class 11 Physics Transmission Of Heat Short Answer Questions

WBCHSE Class 11 Physics Transmission Of Heat

Unit 7 Properties Of Bulk Matter Chapter 9 Transmission Of Heat Short Answer Type Questions

Question 1. At what temperature does a body stop radiating?
Answer:

A body stop radiating at absolute zero or 0 K temperature.

Question 2. The ratio of thermal conductivity of two rods is 4:3. The radii and thermal resistances of the two rods are equal. Calculate the ratio of their length.
Answer:

The ratio of thermal conductivity of two rods is 4:3. The radii and thermal resistances of the two rods are equal.

Thermal resistance = \(\frac{1}{k A} \frac{l}{k}=\frac{1}{\pi r^2}\)

The radii and thermal resistance of the two rods are equal.

∴ \(\frac{1}{k_1} \frac{l_1}{\pi r^2}=\frac{1}{k_2} \frac{l_2}{\pi r^2} \quad \text { or, } \frac{l_1}{l_2}=\frac{k_1}{k_2}=\frac{4}{3}\)

Hence, ratio of their lengths is 4 : 3.

Question 3. The thermal conductivity of aluminum is 0.5 cal • s^-1 • cm^-1 • °C^-1. Express it in SI system.
Answer:

The thermal conductivity of aluminum is 0.5 cal • s^-1 • cm^-1 • °C^-1.

Thermal conductivity of aluminium, k = 0.5 cal • s^-1 • cm^-1 • °C^-1 [In CGS system]

In SI, k = 0.5 x 4.2 x 10² J • m^-1 • K^-1 • s^-1

= 210 J • m^-1 • K^-1 • s^-1

WBCHSE Physics Chapter 9 Solutions 

Question 4. Consider a blackbody radiation In a cubical box at absolute temperature T. If the length of each side of the box is doubled and the temperature of the walls of the box and that of the radiation is halved, then the total energy

1. Halves
2. Doubles
3. Quadruples
4. Remains the same

Answer:

According to Stefan’s law, \(E \propto A \sigma T^4\)

∴ \(\frac{E_1}{E_2}=\frac{A_1 T_1^4}{A_2 T_2^4}=\frac{A_1 T_1^4}{\left(4 A_1\right)\left(\frac{1}{2} T_1^4\right)}\)

because \(A_2=4 A_1 \text { and } T_2=\frac{1}{2} T_1\)

or, \(\frac{E_1}{E_2}=4: 1\)

The option 3 is correct.

WBBSE Class 11 Transmission of Heat Short Answer Questions

Question 5. The same quantity of ice is filled in each of the two metal containers P and Q having the same size, shape, and wall thickness but made of different materials. The containers are kept in identical surroundings. The ice in P melts completely in time t₁ whereas that in Q takes a time t₂. The ratio of thermal conductivities of the materials of P and Q is

1. \(t_2: t_1\)
2. \(t_1: t_2\)
3. \(t_1^2: t_2^2\)
4. \(t_2^2: t_1^2\)

Answer:

⇒ \(\left(k A \frac{d T}{d x}\right) t=m L\),

so, \(k \propto \frac{1}{t}\).

Hence, \(\frac{k_1}{k_2}=\frac{t_2}{t_1}\)

The option 1 is correct.

WBCHSE Class 11 Physics Transmission of Heat 

Question 6. A solid maintained at t₁°C Is kept In an evacuated chamber at temperature t°C(t₂>>t₁). The rate of boat absorbed by the body is proportional to

1. \(t_2^4-t_1^4\)
2. \(\left(t_2^4+273\right)-\left(t_1^4+273\right)\)
3. \(t_2-t_1\)
4. \(t_2^2-t_1^2\)

Answer:

None of the options Is correct.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 7. If the temperature of the sun gets doubled, the rate of energy received on the earth will increase by a factor of

1. 2
2. 4
3. 8
4. 16

Answer:

The rate of energy received on the earth,

E ∝ T⁴ [T = temperature of the sun]

∴ Now, if the temperature of the sun gets doubled, then the rate of energy received on the earth will increase 2⁴ = 16 times.

The option 4 is correct.

Question 8. The temperature of the water of a pond is 0°C while that of the surrounding atmosphere is -20 °C. If the density of ice is ρ, coefficient of thermal conductivity is k and the latent heat of melting is L then the thickness Z of the ice layer formed increases as a function of time as,

1. \(Z^2=\frac{60 k}{\rho L} t\)
2. \(Z=\sqrt{\frac{40 k}{\rho L}} t\)
3. \(Z^2=\frac{40 k}{\rho L} \sqrt{t}\)
4. \(Z^2=\frac{40 k}{\rho L} t\)

Answer:

If the thickness of the ice layer increases by Z in time t, then

t = \(\frac{\rho L}{2 k \theta} Z^2\)

Now, \(\theta =[0-(-20)]=20^{\circ} \mathrm{C}\)

∴ t = \(\frac{\rho L}{2 k \cdot 20} Z^2 \quad \text { or, } Z^2=\frac{40 k t}{\rho L}\)

The option D is correct.

Understanding Heat Transfer Short Answer Questions

Question 9. The temperature of a blackbody radiation enclosed in a container of volume V in increased from 100°C to 1000°C. The heat required in the process is

1. 4.79 x 10^-4 cal
2. 9.21 x 10^-5 cal
3. 2.17 x 10^-4 cal
4. 7.54 x 10^-4 cal

Answer: Data insufficient.

WBCHSE Class 11 Physics Transmission of Heat 

Question 10. Three rods of copper, brass, and steel are welded together to form a Y -shaped structure. Aren of a cross-section of each rod Is 4 cm². The end of the copper rod Is maintained at 100 °C whereas ends of brass and stool are kept at 0 °C. Lengths of the copper, brass, and steel rods are 46,13 and 12 cm respectively. The rods are thermally Insulated from surroundings except at the ends. Thermal conductivities of copper, brass, and steel are 0.92, 0.26, and 0.12 in CGS units, respectively. Rate of heat flow through copper rod is

1. 1.2 cal/s
2. 2.4 calls
3. 4.0 cal/s
4. 6.0 cal/s

Answer:

⇒ \(Q=Q_1+Q_2\)

or, \(\frac{0.92 \times 4(100-T)}{46}\)

= \(\frac{0.26 \times 4 \times(T-0)}{13}+\frac{0.12 \times 4 \times T}{12} \)

or, \(200-2 T=2 T+T\)

or, \(T=40^{\circ} \mathrm{C}\)

Q = \(\frac{0.92 \times 4 \times 60}{46}=4.8 \mathrm{cal} / \mathrm{s}\)

The option 3 is correct.

WBCHSE Physics Chapter 9 Solutions 

Question 11. Consider a spherical shell of radius R at temperature 7. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u = \(\frac{U}{V} \propto T^4\) and pressure \(P=\frac{1}{3}\left(\frac{U}{V}\right)\). If the shell now undergoes an adiabatic expansion the relationship between T and R is

1. \(T \propto e^{-R}\)
2. \(T \propto e^{-3 R}\)
3. \(T \propto \frac{1}{R}\)
4. \(T \propto \frac{1}{R^3}\)

Answer:

For 1 mol ideal gas, \(P V=R_0 T\) [where R₀ is the universal gas constant]

or, \(P=\frac{R_0 T}{V}=\frac{1}{3}\left(\frac{U}{V}\right) \propto T^4\)

So, \(\frac{1}{V} \propto T^3\)

or, \(\frac{1}{\frac{4}{3} \pi R^3} \propto T^3\)

or, \(T^3 \propto \frac{1}{R^3}\)

∴ \(T \propto \frac{1}{R}\)

The option 3 is correct.

Key Concepts in Heat Transfer Short Answers

Question 12. A certain quantity of water cools from 70 °C to 60 °C in the first 5 minutes and to 54 °C in the next 5 minutes. The temperature of the surroundings is

1. 45°C
2. 20°C
3. 42°C
4. 10°C

Answer:

From Newton’s law of cooling, \(\frac{(70+273)-\theta_0}{(60+273)-\theta_0}=\frac{(60+273)-\theta_0}{(54+273)-\theta_0}\)

[θ₀ = temperature of the surrounding]

or, θ₀ = 318 K = 45°C

The option 1 is correct

Question 13. The two ends of a metal rod are maintained at temperatures 100 °C and 110 °C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200 °C and 210 °C, the rate of heat flow will be

1. 44.0 J/s
2. 16.8 J/s
3. 8.0 J/s
4. 4.0 J/s

Answer:

Rate of heat flow is proportional to the difference in temperatures of the two ends.

∴ \(\frac{\left(\frac{d \theta}{d t}\right)_{1 \text { st case }}}{\left(\frac{d \theta}{d t}\right)_{2 \mathrm{nd} \text { case }}}=\frac{\left(\theta_1-\theta_2\right)_{1 \text { st case }}}{\left(\theta_1-\theta_2\right)_{2 \mathrm{nd} \text { case }}}\)

or, \(\left(\frac{d \theta}{d t}\right)_{2 \text { nd case }}=\left(\frac{d \theta}{d t}\right)_{1 \text { st case }} \times \frac{\left(\theta_1-\theta_2\right)_{2 \text { nd case }}}{\left(\theta_1-\theta_2\right){ }_{1 \text { st case }}}\)

= \(4 \times \frac{10}{10}=4 \mathrm{~J} / \mathrm{s}\)

= 4 x 10/10 = 4 J/s

The option 4 is correct.

WBCHSE Physics Chapter 9 Solutions 

Question 14. A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁, at wavelength 500 nm is U_2, and that at 1000 nm is U₃. Wien’s constant b = 2.88 x 10⁶ nm • K. Which of the following is correct?

1. U₃ = 0
2. U₁ >U₂
3. U₂>U₁
4. U₁ = 0

Answer:

According to the Wien’s displacement law, \(\lambda_m T =b\)

or, \(\lambda_m=\frac{b}{T}=\frac{2.88 \times 10^6}{5760}=500 \mathrm{~nm}\)

Now, the maximum amount of emitted radiation corresponding to λ_m is U₂.

From graph, U₁ < U₂ > U₃

∴ U₂ > U₁

The option 3 is correct.

Fourier’s Law of Heat Conduction Short Answers

Class 11 Physics Heat Transfer Questions 

Question 15. In a certain planetary system, it is observed that one of the celestial bodies having a surface temperature of 200 K, emits radiation of maximum intensity near the wavelength 12μm. The surface temperature of a nearby star which emits light of maximum intensity at a wavelength λ = 4800Å, is

1. 7500 K
2. 5000 K
3. 2500 K
4. 10000 K

Answer:

According to the Wien’s displacement law, \(\lambda_m T=\text { constant }\)

∴ \(\lambda_{m_1} T_1=\lambda_{m_2} T_2\)

or, \(T_2=T_1 \frac{\lambda_{m_1}}{\lambda_{m_2}}=200 \times \frac{12 \times 10^{-6}}{4800 \times 10^{-10}}=5000 \mathrm{~K}\)

The option 2 is correct.

Transmission of Heat Class 11 Notes 

Question 16. A wall consists of alternating blocks of length ‘ d ’ and coefficient of thermal conductivity K₁ and K₂ respectively as shown. The cross-sectional area of the blocks are the same. The equivalent coefficient of thermal conductivity of the wall between left and right is

1. \(\frac{K_1+K_2}{2}\)
2. \(\frac{2 K_1 K_2}{K_1+K_2}\)
3. \(\frac{K_1+K_2}{3}\)
4. \(\frac{3 K_1 K_2}{K_1+K_2}\)

Answer:

Thermal resistance of the blocks, \(R_1=R_3=R_5=\frac{1}{K_1} \frac{d}{A} \text { and } R_2=R_4=R_6=\frac{1}{K_2} \frac{d}{A}\)

If R be the equivalent thermal resistance of the parallel combination, then

⇒ \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots=3 \frac{K_1 A}{d}+3 \frac{K_2 A}{d}=\frac{3 A}{d}\left(K_1+K_2\right)\)

If K be the equivalent thermal conductivity of the wall, then

⇒ \(\frac{1}{R}=\frac{6 A}{d} K\)

∴ \(\frac{6 A}{d} K=\frac{3 A}{d}\left(K_1+K_2\right)\)

or, \(K=\frac{K_1+K_2}{2}\)

The option 1 is correct

Question 17. The power radiated by a black body is P and it radiates maximum energy at wavelength, \(\lambda_0\). If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \(\frac{3}{4} \lambda_0\), the power radiated by it becomes nP. The value of n is

1. \(\frac{256}{81}\)
2. \(\frac{4}{3}\)
3. \(\frac{3}{4}\)
4. \(\frac{81}{256}\)

Answer:

From Wien’s displacement law

∴ \(\lambda_{\max } \cdot T=\text { constant }\)

∴ \(\lambda_1 T_1=\lambda_2 T_2 \quad \text { or, } \lambda_0 T_1=\frac{3}{4} \lambda_0 T_2 \quad \text { or, } \frac{T_2}{T_1}=\frac{4}{3}\)

∴ \(\frac{P_2}{P_1}=\left(\frac{T_2}{T_1}\right)^4 \quad \text { or, } \frac{n P}{P}=\left(\frac{4}{3}\right)^4\)

or, \(n=\frac{256}{81}\)

The option 1 is correct.

Heat Transfer MCQs for Class 11 

Real-Life Examples of Heat Transmission

Question 18. Using it find the time taken by a hot food in a pan to cool from 71 °C to 69°C If the room temperature is 20°C. The food cools from 94°C to 86°C in 2 minutes.
Answer:

The time taken by a hot food in a pan to cool from 71 °C to 69°C If the room temperature is 20°C. The food cools from 94°C to 86°C in 2 minutes

Let θ₀ = room temperature; t = time taken by the hot food to cool from θ₁ °C to θ₂ °C Then, from

Newton’s law of cooling, \(k t=\log \frac{\theta_1-\theta_0}{\theta_2-\theta_0}, where k=\mathrm{a} constant\)

In this problem, \(\theta_0=20^{\circ} \mathrm{C}\).

In the hotter region, \(\theta_1=94^{\circ} \mathrm{C}, \theta_2=86^{\circ} \mathrm{C} and t=2 \mathrm{~min}=120 \mathrm{~s}\).

In the colder region, \(\theta_1=71^{\circ} \mathrm{C} and \theta_2=69^{\circ} \mathrm{C}\)

So, \(120 k=\log \frac{94-20}{86-20}=\log \frac{74}{66}=0.0497\)

and \(k t=\log \frac{71-20}{69-20}=\log \frac{51}{49}=0.0174\)

Then, \(\frac{120}{t}=\frac{0.0497}{0.0174}, \quad or, t=\frac{120 \times 0.0174}{0.0497}=42 \mathrm{~s}\)

WBCHSE Physics Chapter 9 Solutions 

Question 19. Show graphically the temperature variation with time associated with a cooling hot body. Why burns from steam are more serious than those from boiling water?
Answer:

Shows the exponentially decreasing nature of temperature with the passage of time.

Every 1 g of steam at 100 °C contains a latent heat excess of about 540 cal, compared to 1 g of water at 100°C.

So, steam releases a greater amount of energy, than that released by boiling water, to a cold body brought to its contact. That is why steam burns are more serious.

Heat Transfer MCQs for Class 11 

Question 20. What is thermal conductivity of perfect heat conductors?
Answer:

Thermal conductivity of perfect heat conductors

Thermal conductivity of a perfect heat conductor is infinity.