Question

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. $16$

Answer 1

Answer: (A)

$6$

Rate of heat loss with only room heater $P_{heater}=\frac{dQ}{dt}=a(16-(-20\left)\right)=a(16+20)=36a$ where a is a constant
Rate of heat loss with room heater and stove together $P_{heater}+P_{stove}=\frac{dQ}{dt}=a(22-(-20\left)\right)=a(22+20)=42a$ where a is a constant

$\therefore \frac{P_{heater}}{P_{heater}+P_{stove}}=\frac{36}{42}=\frac{6}{7}$

$\Rightarrow P_{heater}=6P_{stove}=6\times 1=6kW$ since $P_{stove}=1kW$ as given