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A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 19.0 kW of electric power (generated by doing work at the rate 19.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 19.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 3.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K = QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 250 hours per month during the winter. How much electric power (in kW) does the heat pump use to deliver 19.0 kJ/s of heat energy to the house? Express your answer in kilowatts. ► View Available Hint(s) Win = Submit Part B Cost₁= ——| ΑΣΦ Submit What does one month's heating cost in the home with a 19.0 kW electric heater? Express your answer in dollars. ► View Available Hint(s) ? ΠΙΑΣΦ kW $

A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 19.0 kW of electric power (generated by doing work at the rate 19.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 19.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 3.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K = QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 250 hours per month during the winter. How much electric power (in kW) does the heat pump use to deliver 19.0 kJ/s of heat energy to the house? Express your answer in kilowatts. ► View Available Hint(s) Win = Submit Part B Cost₁= ——| ΑΣΦ Submit What does one month's heating cost in the home with a 19.0 kW electric heater? Express your answer in dollars. ► View Available Hint(s) ? ΠΙΑΣΦ kW $

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 19.0 kW of electric power (generated by doing work at the rate 19.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 19.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 3.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K = QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 250 hours per month during the winter. How much electric power (in kW) does the heat pump use to deliver 19.0 kJ/s of heat energy to the house? Express your answer in kilowatts. ► View Available Hint(s) Win = Submit Part B Cost₁ = 17 ΑΣΦ Submit What does one month's heating cost in the home with a 19.0 kW electric heater? Express your answer in dollars. ► View Available Hint(s) ? VE ΑΣΦ kW ? $
Transcribed Image Text:A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 19.0 kW of electric power (generated by doing work at the rate 19.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 19.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 3.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you get" is heat delivered. So the coefficient of performance of a heat pump is K = QH/Win. An average price for electricity is about 40 MJ per dollar. A furnace or heat pump will run typically 250 hours per month during the winter. How much electric power (in kW) does the heat pump use to deliver 19.0 kJ/s of heat energy to the house? Express your answer in kilowatts. ► View Available Hint(s) Win = Submit Part B Cost₁ = 17 ΑΣΦ Submit What does one month's heating cost in the home with a 19.0 kW electric heater? Express your answer in dollars. ► View Available Hint(s) ? VE ΑΣΦ kW ? $
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