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:

1. Calculate specific heat capacity of water.
2. Heat released in the region BC.

Answer 1

(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 × C_p × (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