A refrigerator removes 250 J of heat from its interior in 60 seconds using input energy of 600 J. What is its efficiency?

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Multiple Choice

A refrigerator removes 250 J of heat from its interior in 60 seconds using input energy of 600 J. What is its efficiency?

A refrigerator’s performance is described by the coefficient of performance (COP), which is the amount of heat removed from the interior (Q_L) per unit of input energy (W_in): COP = Q_L / W_in. Here Q_L is 250 J and W_in is 600 J, so COP = 250 / 600 ≈ 0.4167, which rounds to 0.417. This matches the given data, so it’s the correct value. The other options would require different amounts of heat removed (not the 250 J given) and don’t reflect the actual ratio.

A refrigerator’s performance is described by the coefficient of performance (COP), which is the amount of heat removed from the interior (Q_L) per unit of input energy (W_in): COP = Q_L / W_in. Here Q_L is 250 J and W_in is 600 J, so COP = 250 / 600 ≈ 0.4167, which rounds to 0.417. This matches the given data, so it’s the correct value. The other options would require different amounts of heat removed (not the 250 J given) and don’t reflect the actual ratio.

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