Question

The room heater can maintain only ${{16}^{\circ }}C$ in the room when the temperature outside is $-{{20}^{\circ }}C$. It is not warm and comfortable, that is why the elective stove with power of 1kW is also plugged in. Together these two devices maintain the room temperature of ${{22}^{\circ }}C$. Determine the thermal power of the heater.
$\begin{aligned} & (A)3kW \\\ & (B)4kW \\\ & (C)5kW \\\ & (D)6kW \\\ \end{aligned}$

Answer 1

Apply Newton’s law of cooling. Newton’s law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. By applying this concept and substituting we will get the thermal power of the heater.
Formula used:
$\dfrac{dT}{dt}=k\left( {{T}_{1}}-{{T}_{2}} \right)$
where, $\dfrac{dT}{dt}$ is the rate law of heat loss.
k is a constant
${{T}_{1}}$ and ${{T}_{2}}$ are the temperatures.

Complete answer:
Given that,
$\begin{aligned} & {{T}_{1}}={{16}^{\circ }}C \\\ & {{T}_{2}}=-{{20}^{\circ }}C \\\ \end{aligned}$
Rate of heat loss with room heater,
${{P}_{heater}}=\dfrac{dT}{dt}=k({{T}_{1}}-{{T}_{2}})$ …………….(1)
$\Rightarrow \dfrac{dT}{dt}=k(16-\left( -20 \right))$
$\Rightarrow \dfrac{dT}{dt}=k\left( 16+20 \right)$
$\Rightarrow \dfrac{dT}{dt}=36k$
where, k is a constant.
Rate of heat loss room heater and stove together is,
${{P}_{heater}}+{{P}_{stove}}=\dfrac{dQ}{dt}=k\left( 22-\left( -20 \right) \right)$
$\Rightarrow {{P}_{heater}}+{{P}_{stove}}=k\left( 22+20 \right)$
$\Rightarrow {{P}_{heater}}+{{P}_{stove}}=42k$ …………..(2)
Therefore by dividing equation (1) by equation (2) we get,
$\dfrac{{{P}_{heater}}}{{{P}_{heater}}+{{P}_{stove}}}=\dfrac{36k}{42k}$
$\Rightarrow \dfrac{{{P}_{heater}}}{{{P}_{heater}}+{{P}_{stove}}}=\dfrac{36}{42}=\dfrac{6}{7}$
Rearranging the equation we get,
${{P}_{heater}}=\dfrac{6}{7}\left( {{P}_{heater}}+{{P}_{stove}} \right)$
$\Rightarrow 7{{P}_{heater}}=6\left( {{P}_{heater}}+{{P}_{stove}} \right)$
$\Rightarrow 7{{P}_{heater}}=6{{P}_{heater}}+6{{P}_{stove}}$
$\Rightarrow {{P}_{heater}}=6{{P}_{stove}}$
Given that,
${{P}_{stove}}=1kW$
$\Rightarrow {{P}_{heater}}=6\times 1=6kW$

So, the correct answer is “Option D”.

Additional Information:
According to Newton’s law of cooling, the temperature goes on decreasing with time exponentially. According to Newton’s law of cooling the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings.

Note:
Newton’s law of cooling holds only for very small temperature differences. And the temperature goes on decreasing with the time. The temperature difference is very small and the nature of the heat transfer mechanism is the same. Newton’s law of cooling is followed for force pumped fluid cooling, where the properties of fluid do not vary strongly with temperature. Convection cooling is usually said to be governed by Newton’s law of cooling.