Suppose you must remove an average of 3.9*108J of thermal energy per day to keep your house cool during the summer. Part A If you upgrade from an old air conditioner with a COP of 2.3 to a new air conditioner with a COP of 6.0, by how many joules is the required mechanical work reduced each day?

Question

Engineering

Marin Sawyer

1 May, 07:33

Suppose you must remove an average of 3.9*108J of thermal energy per day to keep your house cool during the summer. Part A If you upgrade from an old air conditioner with a COP of 2.3 to a new air conditioner with a COP of 6.0, by how many joules is the required mechanical work reduced each day?

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Answers (1)

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Frederick Shelton · Answer 1

The required mechanical work is required to reduce each day by 1.05*10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9*10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9*10^8/2.3 = 1.70*10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9*10^8/6 = 6.5*10^7 J

Reduction in mechanical work = (1.7*10^8) - (6.5*10^7) = 1.05*10^8 J