Question

A heater of power 2000 kW is switched on inside a body, so that its surface temperature is maintained at ${{27}^{\circ }}C$. The surrounding temperature is zero kelvin. If the voltage is dropped by 19%, the new equilibrium temperature of the body is (consider heat loss to the surroundings is due to radiation only)

Answer 1

Use the formula for the rate of heat loss of a black body when it is at a constant temperature, higher than the surrounding temperature and write the equation for the rate of heat loss for the given body. Then use the concept that power is directly proportional to voltage.

Formula used:
$\dfrac{dQ}{dt}=\sigma A\left( T_{2}^{4}-T_{1}^{4} \right)$
where $\dfrac{dQ}{dt}$ is the rate of heat loss, $\sigma$ is the stefan’s constant, A is the surface area, ${{T}_{2}}$ is the temperature of the body and ${{T}_{1}}$ is the surrounding temperature.

Complete step by step answer:
It is given that the heater is a power of 2000 kW.
This means that $P=2000kW$.
And this heater is placed inside a body and the temperature of the body is maintained at ${{27}^{\circ }}C$.
This means that ${{T}_{2}}={{27}^{\circ }}C=300K$.
The temperature of the surrounding is given to be at 0K.
This means that ${{T}_{1}}=0K$.
Therefore, the rate of heat loss to the surrounding is equal to,
$\dfrac{dQ}{dt}=\sigma A\left( T_{2}^{4}-T_{1}^{4} \right)$ ….. (i)
The rate of heat loss is equal to the power of the heater, i.e.
$P=\dfrac{dQ}{dt}=2\times {{10}^{6}}W$
Substitute the known values in equation (i).
$\Rightarrow \dfrac{dQ}{dt}=\sigma A\left( {{(300)}^{4}}-0 \right)=2\times {{10}^{6}}$
$\Rightarrow P=\sigma A\left( 81\times {{10}^{8}} \right)=2\times {{10}^{6}}$
$\Rightarrow \sigma A=\dfrac{2\times {{10}^{6}}}{81\times {{10}^{8}}}=\dfrac{2}{81\times {{10}^{2}}}$

Then it is given that the voltage is dropped by 19%. We know that the power of a heater is directly proportional to the square of the voltage. Since the voltage is dropped by 19%, the new rate of heat loss is equal to $P'={{\left( \dfrac{81}{100} \right)}^{2}}P$
$\Rightarrow P'={{\left( \dfrac{81}{100} \right)}^{2}}\times 2\times {{10}^{6}}$
$\Rightarrow P'={{\left( \dfrac{81}{100} \right)}^{2}}\times 2\times {{10}^{6}}=13122\times {{10}^{2}}W$
But we know that $P'=\sigma A\left( {{T}^{4}}-0 \right)$, where T is the new equilibrium temperature.
This means that $\sigma A\left( {{T}^{4}}-0 \right)=13122\times {{10}^{2}}$
Substitute $\sigma A=\dfrac{2}{81\times {{10}^{2}}}$ in the above equation.
Then,
$\Rightarrow \left( \dfrac{2}{81\times {{10}^{2}}} \right)\left( {{T}^{4}}-0 \right)=13122\times {{10}^{2}}$
$\Rightarrow {{T}^{4}}=13122\times {{10}^{2}}\times \dfrac{81\times {{10}^{2}}}{2}$
$\Rightarrow {{T}^{4}}=13122\times {{10}^{2}}\times \dfrac{81\times {{10}^{2}}}{2}\\\ \Rightarrow {{T}^{4}}=531441\times {{10}^{4}}$
$\therefore T=270K$

Therefore, the new equilibrium temperature of the body is 270K.

Note: Many of the students may ignore the fact that the power of a heater is directly proportional to the square of the voltage (potential difference) across the resistor and may make some mistake in solving the given question. You may know this fact from the formula $P=\dfrac{{{V}^{2}}}{R}$, where V is voltage across resistance R.