Q63 / 90 · KVPY 2012 (SB/SX) · Physics
The total energy of a black body radiation source is collected for five minutes and used to heat water. The temperature of the water increases from $10.0^{\circ} \mathrm{C}$ to $11.0^{\circ} \mathrm{C} .$ The absolute temperature of the black body is doubled and its surface area halved and the experiment repeated for the same time. Which of the following statements would be most nearly correct?
- AThe temperature of the water would increase from $10.0^{\circ} \mathrm{C}$ to a final temperature of $12^{\circ} \mathrm{C}$
- Correct answerBThe temperature of the water would increase from $10.0^{\circ} \mathrm{C}$ to a final temperature of $18^{\circ} \mathrm{C}$
- CThe temperature of the water would increase from $10.0^{\circ} \mathrm{C}$ to a final temperature of $14^{\circ} \mathrm{C}$
- DThe iemperuiure of the wuier would increases from $10.0^{\circ} \mathrm{C}$ io u finul iemperaiure of $11^{\circ} \mathrm{C}$
Solution
Energy radiated, $\mathrm{U} \propto \mathrm{AT}^{4} \mathrm{t}$
$\begin{array}{l}
\Rightarrow \frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{A} / 2(2 \mathrm{~T})^{4} \cdot \mathrm{t}}{\mathrm{AT}^{4} \cdot \mathrm{t}}=8 \\
\Rightarrow \mathrm{U}_{2}=8 \mathrm{U}_{1} \\
\Rightarrow \mathrm{mS} \Delta \mathrm{t}_{2}=8 \mathrm{mS} \Delta \mathrm{t}_{1} \\
\Rightarrow \Delta \mathrm{t}_{2}=8 \Delta \mathrm{t}_{1}
\end{array}$
Hence, the correct answer is option: 'B'.