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Kelsie Sawayn

Feb 20, 2025

A freezer has a coefficient of performance of 6.30. The freezer is advertised as using 466 kW-h/yr?

Note: One kilowatt-hour (kWh) is an amount of energy equal to running a 1-kW appliance for one hour. (a) On average, how much energy does it use in a single day? Answer in J (b) On average, how much energy does it remove from the refrigerator in a single day? answer in J(c) What maximum mass of water at 17.1°C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 x 10^5 J/kg, and its specific heat is 4186 J/kg · °C.) answer in kg
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Hermina Ferry

Feb 20, 2025

(a) consumption per day = 466/365 = 1.28 kw-hr(b) By COP(cooling) = Q/W =>Q = W x COP=>Q = 1280 x 6.30 = 8040 J(c) By Q = ms∆t + mL =>8040 = m x 4186 x (100 – 17.1) + m x 333000=>8040 = 680019.4m=>m = 0.01182 kg=>m = 11.83 gm... Show More
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