Question

The room heater can provide only ${16^o}C$ in the room when the temperature outside is $- {20^o}C$ . 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 ${22^o}C$ . Determine the thermal power of the heater.
(A) 3KW
(B) 4KW
(C) 5KW
(D) 6KW

Answer 1

Hint We will first calculate the rate of heat loss when only room heater is working i.e. ${P_{heater}}$ . As we know that power loss is directly proportional to temperature difference.
Then we will calculate the rate of heat loss when room heater and stove both are working i.e. ${P_{heater}} + {P_{stove}}$
Now we will find thermal power of heater in kW using $\dfrac{{{P_{heater}}}}{{{P_{heater}} + {P_{stove}}}}$ . So, after calculating the value of ${P_{heater}}$ we will choose the correct option.

Complete step by step solution
We use a concept for solving this problem i.e. Power loss or rate of heat loss is directly proportional to temperature difference.
By using this we calculate rate of heat loss when only room heater is working i.e. ${P_{heater}}$ .
${P_{heater}} = c({t_h} - {t_s})$ , where ${\text{c}}$ is the constant , ${t_h}$ is the temperature provided by heater and ${t_s}$ is the temperature of surrounding.
So, after substituting the values, ${P_{heater}} = c(16 - ( - 20))$
${P_{heater}} = c(16 + 20) = 36c$
Now rate of heat loss when room heater and stove both are working i.e. ${P_{heater}} + {P_{stove}}$
Now ${P_{heater}} + {P_{stove}} = c({t_{h + s}} - {t_s}) = c(22 - ( - 20)) = c(22 + 20)$ , where ${t_{h + s}}$ is the temperature when room heater and stove both are working ${P_{heater}} + {P_{stove}} = 42c$
Now the thermal power of heater in kW : $\dfrac{{{P_{heater}}}}{{{P_{heater}} + {P_{stove}}}} = \dfrac{{36}}{{42}} = \dfrac{6}{7}$ .
After solving we get,

7{\text{ }}{P_{heater}} = 6({P_{heater}} + {P_{stove}}) \\\ {P_{heater}} = 6{P_{stove}} = 6 \times 1Kw \\\ $$ , since as given $${P_{stove}} = 1Kw$$ . So, we get $${P_{heater}} = 6Kw$$ . **Hence option D is correct.** **Note** Remember that the hotter object transfers its heat to the colder object until the objects are at the same temperature or in other words, they attain thermal equilibrium. The concept of the total heat loss of the object also involves losses occurring by radiation, convection, and conduction. There is no material, which completely prevents heat loss, we can only minimize the heat loss. Remember all the formula of heat loss, rate of heat loss, thermal power etc and also power loss or rate of heat loss is directly proportional to temperature difference.