Heat Transfer - Radiation - NEET Physics Questions
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Heat Transfer - Radiation

Question 1: moderate

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 U2 and that at 1000 nm is U3. Wien’s constant, b=2.88\times 10^{6} nmK. Which of the following is correct?

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

To solve this, we use Wien's Displacement Law, which states:

\[
\lambda_{\text{max}} = \frac{b}{T}
\]

Where:
- \(\lambda_{\text{max}}\) is the wavelength at which the maximum energy is emitted,
- \(b = 2.88 \times 10^6 \, \text{nmK}\) is Wien's constant,
- \(T = 5760 \, K\) is the temperature of the black body.

Step 1: Calculate \(\lambda_{\text{max}}\)
\[
\lambda_{\text{max}} = \frac{2.88 \times 10^6}{5760} = 500 \, \text{nm}
\]

This means that the maximum energy is emitted at a wavelength of 500 nm.

 Step 2: Compare the energies at different wavelengths

- At 500 nm (\( \lambda_{\text{max}} \)): This is where the black body emits the maximum energy. So, \(U_2\) will be the largest.
- At 250 nm: This wavelength is shorter than \(\lambda_{\text{max}}\), and the energy decreases as we move away from the peak, so \(U_1 < U_2\).
- At 1000 nm: This wavelength is longer than \(\lambda_{\text{max}}\), and the energy decreases further, so \(U_3 < U_2\).

Thus, \(U_2 > U_1\), and the correct comparison is \(U_2 > U_1\).

Question 2: moderate

For a black body at temperature 727°C, its rate of energy loss is 20 watt and temperature of surrounding is 227°C. If temperature of black body is changed to 1227°C then its rate of energy loss will be

1. 304 W
2. 320/3 W
3. 240 W
4. 120 W
View Answer

To solve this, we use Stefan-Boltzmann Law for radiative heat transfer:

\[
P = \sigma A (T^4 - T_s^4)
\]

Where:
- \(P\) is the power (rate of energy loss),
- \(\sigma\) is the Stefan-Boltzmann constant,
- \(A\) is the area of the body,
- \(T\) is the temperature of the body in kelvins,
- \(T_s\) is the temperature of the surroundings in kelvins.

 Step 1: Convert temperatures to kelvins
- Initial temperature of the black body: \(727^\circ C = 727 + 273 = 1000\,K\)
- Surrounding temperature: \(227^\circ C = 227 + 273 = 500\,K\)
- New temperature of the black body: \(1227^\circ C = 1227 + 273 = 1500\,K\)

 Step 2: Ratio of power losses
Let the power at \(T_1 = 1000\,K\) be \(P_1 = 20\,W\). The new power at \(T_2 = 1500\,K\) is \(P_2\). Using the Stefan-Boltzmann law:

\[
\frac{P_2}{P_1} = \frac{T_2^4 - T_s^4}{T_1^4 - T_s^4}
\]

Substitute the values:

\[
\frac{P_2}{20} = \frac{1500^4 - 500^4}{1000^4 - 500^4}
\]

Step 3: Simplify the powers of temperatures
\[
\frac{P_2}{20} = \frac{1500^4 - 500^4}{1000^4 - 500^4}
\]

You can approximate the values:

\[
P_2 = 20 \times \frac{(1500^4 - 500^4)}{(1000^4 - 500^4)}
\]

After simplifying, you'll find:

\[
P_2 = \frac{320}{3} \text{ watts}
\]

Thus, the rate of energy loss at 1227°C is \(\frac{320}{3}\,W\).