Class 11 Physics - Transmission of Heat (Important Q&A)
Transmission of Heat Marks · 2M 3M
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📘 2 Mark Q&A
2+ Q&ASolution:
All bodies emit thermal radiation at all temperatures above absolute zero.
According to Stefan–Boltzmann law, the power radiated by a body is proportional to \( T^4 \).
Thus, radiation becomes zero only when \( T = 0 \, \text{K} \).
Since absolute zero (\( 0 \, \text{K} \)) is not practically attainable, a body never completely stops radiating under ordinary conditions.
Therefore, a body stops radiating only at absolute zero temperature (0 K).
Solution:
A perfect heat conductor is a material through which heat flows without any temperature difference across it.
From the relation of heat conduction,
\( Q = \dfrac{K A (T_1 - T_2)}{l} \, t \)
For a perfect conductor, heat can flow even when \( (T_1 - T_2) \to 0 \)
This is possible only if the thermal conductivity \( K \to \infty \).
Therefore, the thermal conductivity of a perfect heat conductor is infinite.
📗 3 Mark Q&A
4+ Q&ASolution:
Let each plate have thickness \( \dfrac{l}{2} \) and equal cross-sectional area \( A \).
Thermal resistance of first plate,
\( R_1 = \dfrac{\frac{l}{2}}{K_1 A} = \dfrac{l}{2K_1 A} \)
Thermal resistance of second plate,
\( R_2 = \dfrac{\frac{l}{2}}{K_2 A} = \dfrac{l}{2K_2 A} \)
Total thermal resistance (since plates are in series),
\( R = R_1 + R_2 = \dfrac{l}{2A} \left( \dfrac{1}{K_1} + \dfrac{1}{K_2} \right) \)
If \( K \) is the equivalent thermal conductivity of the combined plate of thickness \( l \), then
\( R = \dfrac{l}{K A} \)
Therefore,
\( \dfrac{l}{K A} = \dfrac{l}{2A} \left( \dfrac{1}{K_1} + \dfrac{1}{K_2} \right) \)
\( \Rightarrow \dfrac{1}{K} = \dfrac{1}{2} \left( \dfrac{1}{K_1} + \dfrac{1}{K_2} \right) \)
Hence, the equivalent thermal conductivity is
\( K = \dfrac{2K_1K_2}{K_1 + K_2} \)
Solution:
Let the two plates have the same cross-sectional area \( A \) and be maintained at a temperature difference \( \Delta T \).
Heat flows through the plates one after another (in series).
Thermal resistance of first plate,
\( R_1 = \dfrac{d_1}{K_1 A} \)
Thermal resistance of second plate,
\( R_2 = \dfrac{d_2}{K_2 A} \)
Total thermal resistance of the combined system,
\( R = R_1 + R_2 = \dfrac{d_1}{K_1 A} + \dfrac{d_2}{K_2 A} \)
Let \( K \) be the equivalent thermal conductivity of a single plate of thickness \( d_1 + d_2 \)
Then its thermal resistance is
\( R = \dfrac{d_1 + d_2}{K A} \)
Equating the two expressions of \( R \), we get
\( \dfrac{d_1 + d_2}{K A} = \dfrac{d_1}{K_1 A} + \dfrac{d_2}{K_2 A} \)
\( \Rightarrow \dfrac{d_1 + d_2}{K} = \dfrac{d_1}{K_1} + \dfrac{d_2}{K_2} \)
Therefore,
\( K = \dfrac{d_1 + d_2}{\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}} \)
Hence proved.
Solution:
Stefan’s Law: The total energy radiated per second by a black body is directly proportional to its surface area and the fourth power of its absolute temperature.
Mathematically, \( E = \sigma A T^4 \)
For a sphere, surface area \( A = 4\pi r^2 \)
Thus, energy radiated per second,
\( E \propto r^2 T^4 \)
Therefore,
\( \dfrac{E_1}{E_2} = \dfrac{r_1^2 T_1^4}{r_2^2 T_2^4} \)
Given: \( r_1 = 1 \text{ m}, \; r_2 = 4 \text{ m} \)
\( \Rightarrow T_1 = 4000 \text{ K}, \; T_2 = 2000 \text{ K} \)
\( \Rightarrow \dfrac{E_1}{E_2} = \dfrac{1^2 \times (4000)^4}{4^2 \times (2000)^4} \)
\( = \dfrac{1 \times (2 \times 2000)^4}{16 \times (2000)^4} \)
\( = \dfrac{16 (2000)^4}{16 (2000)^4} \)
\( = 1 \)
Therefore, the ratio of energy radiated per second is
\( E_1 : E_2 = 1 : 1 \)
Solution:
Newton’s Law of Cooling: The rate of loss of heat (or rate of fall of temperature) of a body is directly proportional to the difference between its temperature and that of the surroundings, provided the temperature difference is small.
Mathematically,
\( -\dfrac{dT}{dt} \propto (T - T_s) \)
\( \therefore -\dfrac{dT}{dt} = k (T - T_s) \)
where \( T \) is the temperature of the body, \( T_s \) is the temperature of the surroundings, and \( k \) is a constant.
Graph: The temperature variation with time associated with a cooling hot body.
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