Explanation Write the expression for COP of electric resistance for removing heat. COP rev = 1 1 − T L T H (I) Here, reversible coefficient of performance is COP rev , outside temperature is T L and inside temperature of house is T H . Calculate the reversible work input. W ˙ rev,in = Q ˙ H COP rev (II) Here, reversible power input is W ˙ rev,in and heat rate is Q ˙ H . Calculate the irreversibility. I ˙ = W ˙ useful − W ˙ rev,in (III) Here, irreversibility is I ˙ and useful power is W ˙ useful . The heat loss rate from the house is equal to the useful work to remove that heat from the house. Write the expression to calculate heat loss rate from the house. W ˙ useful = Q ˙ H (IV) Conclusion: Substitute 4 ° C for T L and 25 ° C for T H in Equation (I). COP rev = 1 1 − ( 4 + 273.15 ) K ( 25 + 273.15 ) K = 1 1 − 277

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Chapter 8.8, Problem 21P

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The reversible work input for electric resistance heater.

The irreversibility of electric resistance heater.

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Chapter 8.8, Problem 20P

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The reversible work input for electric resistance heater. The irreversibility of electric resistance heater.