Question

The total energy radiated from a black body source is collected for $1 \, min$ and is used to heat a quantity of water. The temperature of water is found to increase from $20 \, ℃$ to $20.5 \, ℃$ . If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20 \, ℃$ , the temperature of water will be

• A. $21^\circ C$
• B. $22 \, ^\circ C$
• C. $24 \, ^\circ C$
• D. $28 \, ^\circ C$

Answer 1

Answer: (D)

From Stefan's law of radiation,
$E \propto T^{4} \, \Rightarrow \, \frac{E_{1}}{E_{2}}=\frac{T_{1}^{4}}{T_{2}^{4}}$
Given, $T_{1}=T, \, \, \, T_{2}=2T$
$\therefore $ $\frac{E_{1}}{E_{2}}=\frac{\left(T\right)^{4}}{\left(2 T\right)^{4}}=\frac{1}{2^{4}}=\frac{1}{16}$
$E_{2}=16 \, E_{1}$
Heat taken by water from radiation
$E=mc\Delta \theta $
Where c is specific heat, $\Delta \theta $ the change in temperature and $m$ the mass.
$\therefore $ $E=m\times 1\times \left(\right.20.5-20\left.\right)$
$E=m\times 0.5$ ...(i)
When energy supplied is 16 times the previous one, then let the temperature rise to $\theta ′$
$\therefore \, \, 16E=m\times 1\times \left(\right.\left(\theta \right)^{′}-20\left.\right)$ ...(ii)
Dividing Eq. (i) by (ii), we get
$\frac{1}{16}=\frac{0.5}{\theta ^{′} - 20}$
$\Rightarrow \, \, \theta ^{′}-20=16\times 0.5=8$
$\Rightarrow \, \, \theta ^{′}=20+8=28℃$