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.

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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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
kg ) (4 186 J/kg · °C)
x 10' J.
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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 kg ) (4 186 J/kg · °C) x 10' J. Submit Skip (you cannot come back)
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