80g
of warm water at 60°C is poured into a calorimeter
containing 100 g of cold water at 10°C. If the final
temperature of mixture is 28°C, find the heat capacity
of the calorimeter. Given that specific heat capacity of
water is 4200 J kg-1 K-1.
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Heat lost by water to change it is temperature from 60℃ to 30℃=ms△T.
=40×4.2×30
=5040J.
Heat gained by water,
=50×4.2×10
=2100J.
Heat gained by vessel,
= Heat capacity ×10
Heat lost by hot water,
= heat lost by cold water + heat gained by vessel.
∴ Heat gained by vessel,
=5040−2100
=2940.
Heat capacity×10=2940
Heat capacity =2940/10
=294J/℃
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