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Question
Calculate the amount of heat released when 5.0 g of water at 20°C is changed into ice at 0°C.
(Specific heat capacity of water = 4.2 J/g°C
Specific latent heat of fusion of ice = 336 J/g)
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Solution
5g water at 20° C \[\underrightarrow{mcθ}\] 5g water at 0° C \[\underrightarrow{mL}\] 5 g ice at 0°C.
∴ Amount of heat released = mcθ + mL
= 5 × 4.2 × (20° - 0°) + 5 × 336 = 21.0 × 20 + 1680
= 420 + 1680 = 2100 Joules
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