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Question
The diagram below shows a cooling curve for 200 g of water. The heat is extracted at the rate of 100 Js-1. Answer the questions that follow:
- Calculate specific heat capacity of water.
- Heat released in the region BC.
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Solution
(a) mass of water (m) = 200 g = 0.2 kg.
Rate of heat extraction= 100 J/s.
So, total heat extracted for curve AB is = 100 × 640 = 64000 J
Now, for curve AB we can write,
64000 = 0.2 × Cp × (353 - 273) ...`[(80^circ "C" = 273 + 80 = 353;),(0^circ "C" = 273 + 0 = 273 "K")]`
Solving we get,
C = 4000 Jkg-1 K-1
Thus, the specific heat capacity of water is 4000 Jkg-1 K-1.
(b) From the figure it is clear that at point Band C is corresponds to time 640 sand 1312 s, respectively. Thus, heat released in region BC is = 100 × (1312 - 640) = 67,200 J
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