Question

The total energy radiated from a black body source is collected for one minute and is used to heat a quantity of water. The temperature of water is found to increases from $20^\circ {\text{C}}$ to $20.5^\circ {\text{C}}$. If the absolute temperature of black body is doubled and the experiment is repeated with the same quantity of water at $20^\circ {\text{C}}$, the temperature of water will be:
A. $28^\circ {\text{C}}$
B. $22^\circ {\text{C}}$
C. $24^\circ {\text{C}}$
D. $21^\circ {\text{C}}$

Answer 1

Use the formula for Stefan-Boltzmann law and derive the relation between heats radiated by black body for its two surface temperatures. Then use the formula for heat exchanged by an object with the surrounding in terms of its mass, specific heat and change in temperature and determine the increased temperature of the water.

Formulae used:
The expression for Stefan-Boltzmann law is
$Q = \sigma {T^4}$ …… (1)
Here, $Q$ is the energy (in the form of heat) radiated by the back body$\sigma$is Stefan’s constant and $T$ is the temperature of the surface temperature of the black body.
The heat exchanged $Q$ by an object with the surrounding is given by
$Q = ms\Delta T$ …… (2)
Here, $m$ is the mass of the object, $s$ is specific heat of the object and $\Delta T$ is the change in temperature of the object.

Complete step by step answer:
We have given that the temperature of the water increases from to when it is heated with the heat radiated ${Q_1}$ by the black body when the surface temperature of the black body is $T$.
${T_1} = T$

Let ${Q_2}$ be the heat radiated by the black body when its surface temperature is $2T$.
${T_2} = 2T$
From equation (1), we can conclude that the energy radiated by the black body is directly proportional to fourth power of surface temperature of the black body.
$Q \propto {T^4}$
Rewrite the above relation for above two values of heat radiated by the black body.
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{T_1^4}}{{T_2^4}}$
Substitute $T$ for ${T_1}$ and $2T$ for ${T_2}$ in the above equation.
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T^4}}}{{{{\left( {2T} \right)}^4}}}$
$\Rightarrow \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T^4}}}{{16{T^4}}}$
$\Rightarrow {Q_2} = 16{Q_1}$

The change in temperature of the water for the heat ${Q_1}$ of black body is
$\Delta T = 20.5^\circ {\text{C}} - 20^\circ {\text{C}}$
$\Rightarrow \Delta T = 0.5^\circ {\text{C}}$
Rewrite equation (2) for the heat ${Q_1}$ exchanged with water.
${Q_1} = ms\left( {0.5^\circ {\text{C}}} \right)$ …… (3)
Here, $m$ is the mass of the water that is heated and $s$ is specific heat of the water.
Rewrite equation (2) for the heat ${Q_2}$ exchanged with water.
${Q_2} = ms\left( {{T_w} - 20^\circ {\text{C}}} \right)$ …… (4)
Here, ${T_w}$ is the increased temperature of the water.

Divide equation (3) by equation (4).
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{ms\left( {0.5^\circ {\text{C}}} \right)}}{{ms\left( {{T_w} - 20^\circ {\text{C}}} \right)}}$
$\Rightarrow \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{0.5^\circ {\text{C}}}}{{{T_w} - 20^\circ {\text{C}}}}$
Substitute $16{Q_1}$ for ${Q_2}$ in the above equation.
$\Rightarrow \dfrac{{{Q_1}}}{{16{Q_1}}} = \dfrac{{0.5^\circ {\text{C}}}}{{{T_w} - 20^\circ {\text{C}}}}$
$\Rightarrow \dfrac{1}{{16}} = \dfrac{{0.5^\circ {\text{C}}}}{{{T_w} - 20^\circ {\text{C}}}}$
$\Rightarrow {T_w} - 20^\circ {\text{C}} = 16 \times 0.5^\circ {\text{C}}$
$\Rightarrow {T_w} = 8 + 20^\circ {\text{C}}$
$\therefore {T_w} = 28^\circ {\text{C}}$
Therefore, the temperature of water is $28^\circ {\text{C}}$.

Hence, the correct option is A.

Note: The students should remember that the mass of the water used in both the cases is the same. Hence, there is the same mass of water for both the cases of heat exchange. Also the specific heat of water is a constant quantity. Hence, it will also be the same for both the cases of heat exchange.