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  • 4 months ago

    To heat 1 kg (1.00 liter) of water from 283.15 K to 303.15 K (10 °C to 30 °C) requires 83.6 kJ. However, to melt ice also requires energy. We can treat these two processes independently; thus, to heat 1 kg of ice from 273.15 K to water at 293.15 K (0 °C to 20 °C) requires:

    (1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt

    PLUS

    (2) 4.18 J/(g·K) × 20K = 4.18 kJ/(kg·K) × 20K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K

    = 417.15 kJ

    From these figures it can be seen that one part ice at 0 °C will cool almost exactly 4 parts water from 20 °C to 0 °C.

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