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Chapter 6: Problem 6

In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped only with two \(1,500 \mathrm{~W}\) space heaters, what is the coldest it can get outside if the house is to maintain an internal temperature of \(20^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The coldest possible outside temperature is 5°C.

Step by step solution

01

Calculate total heat provided by space heaters

First, let's determine the total heat being provided by the two space heaters. Each space heater has a power of 1500 W, so we can find the sum of their powers: $$Total\_heat = 1500 \mathrm{~W}+ 1500 \mathrm{~W}$$
02

Calculate the maximum heat loss

Now we need to find the maximum heat loss the house can have while still maintaining an internal temperature of 20°C. We know that the heat loss rate is 200 W/°C, therefore maximum heat loss is equal to the total heat provided by the space heaters: $$Maximum\_heat\_loss = Total\_heat$$
03

Calculate the maximum temperature difference between inside and outside

Next, we'll need to find the maximum temperature difference between the inside and outside of the house. We can do this by dividing the maximum heat loss by the heat loss rate: $$Maximum\_temperature\_difference = \frac{Maximum\_heat\_loss}{Heat\_loss\_rate}$$
04

Determine the coldest possible outside temperature

Finally, we can determine the coldest possible outside temperature by subtracting the maximum temperature difference from the internal temperature: $$Coldest\_outside\_temperature = Internal\_temperature - Maximum\_temperature\_difference$$ Now let's use these steps to solve the problem.
05

Calculate total heat provided by space heaters

$$Total\_heat = 1500 \mathrm{~W}+ 1500 \mathrm{~W} = 3000 \mathrm{~W}$$
06

Calculate the maximum heat loss

$$Maximum\_heat\_loss = Total\_heat = 3000 \mathrm{~W}$$
07

Calculate the maximum temperature difference between inside and outside

$$Maximum\_temperature\_difference = \frac{Maximum\_heat\_loss}{Heat\_loss\_rate} = \frac{3000 \mathrm{~W}}{200 \mathrm{~W}/^{\circ}\mathrm{C}} = 15^{\circ}\mathrm{C}$$
08

Determine the coldest possible outside temperature

$$Coldest\_outside\_temperature = Internal\_temperature - Maximum\_temperature\_difference = 20^{\circ}\mathrm{C} - 15^{\circ}\mathrm{C} = 5^{\circ}\mathrm{C}$$ Therefore, the coldest it can get outside while still maintaining an internal temperature of 20°C is 5°C.

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Most popular questions from this chapter

Provide at least one example not listed in the text in which heat flows into some other form of energy. \(^{66}\) In the text, we mentioned hot air over a car, wind, internal combustion, and a steam turbine plant.

In a house achieving a heat loss rate of \(200 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) equipped a \(5,000 \mathrm{~W}\) heater, what will the internal temperature be if the outside temperature is \(-10^{\circ} \mathrm{C}\) and the heater is running \(100 \%\) of the time?

What would the maximum thermodynamic efficiency be of some heat engine operating between your skin temperature and the ambient environment \(20^{\circ} \mathrm{C}\) cooler than your skin?

What is the only form of significant electricity production in the U.S. that does not involve a spinning shaft?

How much will it cost per day to keep a house at \(20^{\circ} \mathrm{C}\) inside when the external temperature is steady at \(-5^{\circ} \mathrm{C}\) using direct electric heating \(^{65}\) if the house is rated at \(150 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) and electricity costs \(\$ 0.15 / \mathrm{kWh} ?\)
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