It is known that the temperature in the room is $+ 20^{0} C$ when the outdoor temperature is $- 20^{0} C a n d + 10^{0} C$ when the outdoor temperature is $- 40^{0} C$ . Then the temperature T of the radiator heating the room is (Assuming that radiated by the heater is proportional to the temperature difference with the room)

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$40^{0} C$

$60^{0} C$

$30^{0} C$

$20^{0} C$

answer is B.

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Detailed Solution

We must take into account here that the heat transferred per unit time is proportional to the temperature difference. Let us introduce the following notation : $T_{o u t 1}, T_{o u t 2}$ and $T_{r 1}, T_{r 2}$ are the temperatures outdoors and in the room in the first and second cases respectively. The thermal power dissipated by the radiator in the room is $k_{1} (T - T_{r})$ , where $k_{1}$ is a certain coefficient. The thermal power dissipated from the room is $k_{2} (T_{r} - T_{o u t})$ , where $k_{2}$ is another coefficient. In thermal equilibrium, the power dissipated by the radiator is equal to the power dissipated from the room. Therefore, we can write
$k_{1} (T - T_{r 1}) = k_{2} (T_{r 1} - T_{o u t 1})$
Similarly, in the second case,
$k_{1} (T - T_{r 2}) = k_{2} (T_{r 2} - T_{o u t 2})$
Dividing the first equation by the second, we obtain
$\frac{T - T_{r 1}}{T - T_{r 2}} = \frac{T_{r 1} - T_{o u t 1}}{T_{r 2} - T_{o u t 2}}$
Hence we can determine T:
$T = \frac{T_{r 2} T_{o u t 1} - T_{r 1} T_{o u t 2}}{T_{r 2} + T_{o u t 1} - T_{o u t 2} - T_{r 1}} = 60^{0} C$