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# 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

Last updated date: 01st Mar 2024
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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.

The rate of heat loss with only room heater ${P_{heater}}$= $\dfrac{{dQ}}{{dt}} = a\left( {16 - \left( { - 20} \right)} \right) = a\left( {16 + 20} \right) = 36a$
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$\;\;{P_{stove}} = 1kW\;as\;given.$