Answer

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**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.

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