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

1 Answer

  • (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

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