Question

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 ${10.0}^{o}C$ to ${11.0}^{o}C$. 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 ?

• A. The temperature of the water would increase from ${10.0}^{o}C$ to a final temperature of ${12}^{o}C$
• B. The temperature of the water would increase from ${10.0}^{o}C$ to a final temperature of ${18}^{o}C$
• C. The temperature of the water would increase from ${10.0}^{o}C$ to a final temperature of ${14}^{o}C$
• D. The temperature of the water would increase from ${10.0}^{o}C$ to a final temperature of ${11}^{o}C$

Answer 1

Answer: (B)

The temperature of the water would increase from ${10.0}^{o}C$ to a final temperature of ${18}^{o}C$

By the Stefan's Law of Radiation, the energy emitted from a black body at a temperature $T$ and a surface area $A$, for a time $\Delta t$ is

$\Delta Q=\sigma A{T}^4\Delta t$

It is given that,

$A_2=0.5\times A_1$

$T_2=2\times T_1$

$\Delta Q_2=\sigma A_2{T}_2^4=\sigma (0.5\times A_1)(2\times T_1{)}^4=8\sigma A_1{T}_1^4=8\Delta Q_1$

Let the rise in temperature of the water be $\delta T$, mass of water be $m$ and specific heat be $s$

We then have,

$\Delta Q_1=ms\times 1^{o}C$ ----------(1)

$\Delta Q_2=ms\times \delta T$ ----------(2)

Taking ratio of (1) and (2), we get

$\frac{\delta T}{1^{o}C}=8\Rightarrow \delta T=8^{o}C$

Thus, the temperature of water raises from ${10}^{o}C$ to ${18}^{o}C$