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A well-insulated electric water heater warms 116 kg of water from 20.0° C to 42.0° C in 24.0 min. Find the resistance of its heating element, which is connected across a 240-V potential difference. Part 1 of 5 - Conceptualize We expect the water heater to be high-powered compared to other household appliances. For the power to be on the order of kilowatts, the current would be several amperes and the resistance only a few ohms. Part 2 of 5 - Categorize We will find the energy required to heat the water. Then we will find the power of the heating element and calculate its resistance from the power and the potential difference. Part 3 of 5 - Analyze We know the mass of water m and the temperature change AT. The specific heat of water is c = 4 186 J/kg · °C. For the energy input by heat, we have Q = mcAT 116 116 kg) (4 186 J/kg · °C) 22 22 °C = 1.068 1.07 x 107J. Part 4 of 5 - Analyze If the tank is well-insulated, essentially all of the energy electrically transmitted to the heating element becomes internal energy in the water and we can equate the energy input by heat to the electrical energy transmitted, that is, Q = E .Thus, the power supplied by the heater is elec E elec = P = At Δt 1.07 1.07 x 10' J) 1440 1440 s 7430.55 7420 W. Part 5 of 5 - Analyze The resistance is given by the following. (A)? R = P

A well-insulated electric water heater warms 116 kg of water from 20.0° C to 42.0° C in 24.0 min. Find the resistance of its heating element, which is connected across a 240-V potential difference. Part 1 of 5 - Conceptualize We expect the water heater to be high-powered compared to other household appliances. For the power to be on the order of kilowatts, the current would be several amperes and the resistance only a few ohms. Part 2 of 5 - Categorize We will find the energy required to heat the water. Then we will find the power of the heating element and calculate its resistance from the power and the potential difference. Part 3 of 5 - Analyze We know the mass of water m and the temperature change AT. The specific heat of water is c = 4 186 J/kg · °C. For the energy input by heat, we have Q = mcAT 116 116 kg) (4 186 J/kg · °C) 22 22 °C = 1.068 1.07 x 107J. Part 4 of 5 - Analyze If the tank is well-insulated, essentially all of the energy electrically transmitted to the heating element becomes internal energy in the water and we can equate the energy input by heat to the electrical energy transmitted, that is, Q = E .Thus, the power supplied by the heater is elec E elec = P = At Δt 1.07 1.07 x 10' J) 1440 1440 s 7430.55 7420 W. Part 5 of 5 - Analyze The resistance is given by the following. (A)? R = P

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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Tutorial Exercise A well-insulated electric water heater warms 116 kg of water from 20.0° C to 42.0° C in 24.0 min. Find the resistance of its heating element, which is connected across a 240-V potential difference. Part 1 of 5 - Conceptualize We expect the water heater to be high-powered compared to other household appliances. For the power to be on the order of kilowatts, the current would be several amperes and the resistance only a few ohms. Part 2 of 5 - Categorize We will find the energy required to heat the water. Then we will find the power of the heating element and calculate its resistance from the power and the potential difference. Part 3 of 5 - Analyze We know the mass of water m and the temperature change AT. The specific heat of water is C = 4 186 J/kg · °C. For the energy input by heat, we have Q = mcAT 116 116 kg) (4 186 J/kg · °C) (|22| 22 °c) = 1.068 1.07 x 107 J. Part 4 of 5 - Analyze If the tank is well-insulated, essentially all of the energy electrically transmitted to the heating element becomes internal energy in the water and we can equate the energy input by heat to the electrical energy transmitted, that is, Q = E .Thus, the power supplied by the heater is elec" E elec P = At At 1.07 V 1.07 x 10' J) |1440 1440 s 7430.55 7420 w. Part 5 of 5 - Analyze The resistance is given by the following. (A)? R =
Transcribed Image Text:Tutorial Exercise A well-insulated electric water heater warms 116 kg of water from 20.0° C to 42.0° C in 24.0 min. Find the resistance of its heating element, which is connected across a 240-V potential difference. Part 1 of 5 - Conceptualize We expect the water heater to be high-powered compared to other household appliances. For the power to be on the order of kilowatts, the current would be several amperes and the resistance only a few ohms. Part 2 of 5 - Categorize We will find the energy required to heat the water. Then we will find the power of the heating element and calculate its resistance from the power and the potential difference. Part 3 of 5 - Analyze We know the mass of water m and the temperature change AT. The specific heat of water is C = 4 186 J/kg · °C. For the energy input by heat, we have Q = mcAT 116 116 kg) (4 186 J/kg · °C) (|22| 22 °c) = 1.068 1.07 x 107 J. Part 4 of 5 - Analyze If the tank is well-insulated, essentially all of the energy electrically transmitted to the heating element becomes internal energy in the water and we can equate the energy input by heat to the electrical energy transmitted, that is, Q = E .Thus, the power supplied by the heater is elec" E elec P = At At 1.07 V 1.07 x 10' J) |1440 1440 s 7430.55 7420 w. Part 5 of 5 - Analyze The resistance is given by the following. (A)? R =
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