The room heater can maintain only $16^\circ C$ temperature in the room when the temperature of outside is $- 20^\circ C$. It is not warm and comfortable that is why the electric stove with power of 1kW is also plugged in. Together these two devices maintain the room temperature of $22^\circ C$. Find thermal power of heater in kW:
(A) 6
(B) 12
(C) 18
(D) 16

Hint It should be known to us that the high pressure heaters are used to prevent the Boiler feed water. Moreover they work on the principle of heat exchange. Using this concept we can solve this question.

Complete step by step answer
We know that:
The rate of heat loss with only room heater ${P_{heater}}$= $\dfrac{{dQ}}{{dt}} = a\left( {16 - \left( { - 20} \right)} \right) = a\left( {16 + 20} \right) = 36a$
Where a is constant.
The rate of heat loss with room heater and stove together ${P_{heater}}$+${P_{stove}}$$\dfrac{{dQ}}{{dt}} = a\left( {20 - \left( { - 20} \right)} \right) = a\left( {22 + 20} \right) = 42a$ where a is a constant.
$\dfrac{{{P_{heater}}}}{{{P_{heater}} + {P_{stove}}}} = \dfrac{{36}}{{42}} = \dfrac{6}{7}\;$
$\Rightarrow \;{P_{heater}} = {P_{stove}} = 6 \times 1 = 6kW$
Since$\;\;{P_{stove}} = 1kW\;as\;given.$

So, the correct answer is Option A.

Note We should know that in case of a feed water heater which is a power plant the component used to pre heat the water that is delivered to the stream which generates the boiler. This helps in the reduction of the plant operating costs and also helps to avoid the thermal shock to the boiler metal when the feed water is introduced back into the stream cycle.