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24 October, 18:03

how much energy is required to turn 20 gram cube of ice at a temperature of - 10 degrees Celsius into liquid water at a temperature of 10 degrees Celsius

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  1. 24 October, 19:37
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    Given that:

    Mass (m) = 20 g = 0.02 Kg,

    temperature (T₁) = - 10°C = - 10+273 = 263 K

    temperature (T₂) = 10°C = 10+273 = 283 K

    Specific heat of water (Cp) = 4.187 KJ/Kg k

    We know that Heat transfer (Q) = m. Cp. (T₂ - T₁)

    = 0.02 * 4.187 * (283-263)

    Q = 1.67 KJ

    Heat transferred is 1.67 KJ
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