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Heat Pump. A heat pump is a heat engine run in reverse. In

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 42P Chapter 20

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 42P

Heat Pump?. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In sutuner it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is ?5.0°C and the inside temperature is 17.0°C, how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? (b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at 68°F. Show that the heat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.

Step-by-Step Solution:

Solution 42P Introduction We will use the formula of coefficient of performance to calculate the energy supplied to the room for each jule of electrical energy used. Then we will use the energy output of the resistor to compare which one will deliver more energy to the room. And at last we have to show that if the energy outside decreases, the heat supplied to the room by the hat pump will decreases for same amount of supplied electrical energy. Step 1 The coefficient of performance of a heat pump is given by ….. (1) Here W is the supplied electrical energy and C is the energy released by the pump to the room.cTand T is the temperature of hot and cold air. h So we have W = 1 J T c 5°C = 268 K T h 17°C = 290 K So we can write So for each joule of electrical energy supplied to the pump, it will deliver 12.2 J of heat to the room. Step 2 Since the electrical resistance convert electrical energy to heat. We can not get more than the supplied energy. Hence to receive 12.2 J of energy from electrical resistance, in an ideal condition, we have to supply 12.2 J of electrical energy.

Step 3 of 3

Chapter 20, Problem 42P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “Heat Pump?. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In sutuner it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is ?5.0°C and the inside temperature is 17.0°C, how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? (b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at 68°F. Show that the heat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.” is broken down into a number of easy to follow steps, and 215 words. The full step-by-step solution to problem: 42P from chapter: 20 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 42P from 20 chapter was answered, more than 247 students have viewed the full step-by-step answer. This full solution covers the following key subjects: heat, Pump, inside, air, outside. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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