A freezer has a coefficient of performance of 6.30. The freezer is advertised as

Question

A freezer has a coefficient of performance of 6.30. The freezer is advertised as using 515 kW-h/y. Note: One kilowatt-hour (kW-h) is an amount of energy equal to operating a 1-kW appliance for one hour. (a) On average, how much energy does the freezer use in a single day? J (b) On average, how much thermal energy is removed from the freezer each day? J (c) What maximum amount of water at 19.0

Explanation / Answer

a> 515/365 * 3.6*10^6 = 5.08 *10^6 J per day b> H = 6.3*5.08*10^6 = 32*10^6 J pr day i know this much .. hope it helps :)

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