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Chapter 11: Problem 9

It is known that the temperature in the room is \(+20^{\circ} \mathrm{C}\) when the outdoor temperature is \(-20^{\circ} \mathrm{C}\) and \(+10^{\circ} \mathrm{C}\) when the outdoor temperature is \(-40^{\circ} \mathrm{C}\). Then what is the temperature \(T\) of the radiator heating the room? (Assuming that radiated by the heater is proportional to the temperature difference with the room.) (A) \(40^{\circ} \mathrm{C}\) (B) \(60^{\circ} \mathrm{C}\) (C) \(30^{\circ} \mathrm{C}\) (D) \(20^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature of the radiator is (D) \(20^{\circ} \mathrm{C}\).

Step by step solution

01

Set up the equations based on the given conditions

\ Let R be the room temperature and O be the outdoor temperature. According to the problem, the heat radiated by the heater is proportional to the temperature difference between the room and the heater. So, we can write: \(H = k(T - R)\) Where H is the heat radiated, T is the temperature of the radiator, R is the room temperature, and k is a constant of proportionality. Now we have two conditions that can be used to write two equations: Condition 1: Room temperature is +20掳C when outdoor temperature is -20掳C \(H_1 = k(T - 20)\) \(O_1 = -20掳C\) Condition 2: Room temperature is +10掳C when outdoor temperature is -40掳C \(H_2 = k(T - 10)\) \(O_2 = -40掳C\)
02

Determine the heat ratio from the conditions

\ As the heat radiated is proportional to the temperature difference with the room, we can express the ratio of the two heats as: \(\frac{H_1}{H_2} = \frac{k(T-20)}{k(T-10)}\) Simplifying the equation, we get: \(\frac{T-20}{T-10} = \frac{H_1}{H_2}\)
03

Express the heat as a function of outdoor temperature

\ With the given outdoor temperatures, we can express the heats H_1 and H_2 as functions of the outdoor temperatures: \(H_1 = k(T - R_1) = k(T - 20)\) \(H_2 = k(T - R_2) = k(T - 10)\)
04

Use outdoor temperatures to find the heat ratio

\ Using the outdoor temperatures as given, the ratio of the heats can be rewritten as: \(\frac{(T - 20)}{(T - 10)} = \frac{-20-(-20)}{-40-(-20)}\) Simplifying the equation, we get: \(\frac{(T - 20)}{(T - 10)} = \frac{0}{-20}\) Since the heat ratio is zero, one heat must be zero.
05

Solve for the temperature of the radiator

\ As the heat ratio is zero, we can conclude H_1 = 0 which means the radiator temperature does not change in the first condition. Therefore: \(0 = k(T - 20)\) Solving for T: \(T = 20^{\circ} \mathrm{C}\) So the temperature of the radiator is: (D) \(20^{\circ} \mathrm{C}\)

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