Question

The room heater can provide only ${16}^{\circ }\text{C}$ in the room when the temperature outside is $-{20}^{\circ }\text{C}$. It is not warm and comfortable, that is why the electric stove with power of $1\text{kW}$ is also plugged in. Together these two devices maintain the room temperature of ${22}^{\circ }\text{C}$. The thermal power of the heater (in kW) is

Answer 1

$\Delta T=T_b-T_s=16-(-20)=36K$

For small temperature differences between a body and surroundings, Newton's law of cooling can be applied.

$\frac{dT}{dt}\propto \Delta T$

Multiplying both sides of eqn by $ms$,

$ms\frac{dT}{dt}\propto ms\Delta T$

Heat lost by the room due to radiation
$\frac{dQ}{dt}\propto \Delta T$

Since the room is maintained at constant temperature, the power supplied should equal heat loss due to radiation.

In first case, power supplied by heater
$P\propto 36$

In second case, power supplied by heater and stove
$P+1\propto 42$

The proportionality constant is a property of the room and hence constant.
So, $P=C.36$
$P+1=C.42$

Solving the two equations, $P=6kW$