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Question

A 136.4-g of water at 21.1 C is heated to steam at 125.4 C. How much heat was absorbed?

Answer 1

Specific heat of water = 0.00418 kJ/g/ºC

1. To heat the water from 21.1 ºC to 100 ºC

= 78.9 x 0.00418 = 0.33 kJ/g

= 136.4 g x 0.33 = 44.98 kJ

Latent heat of vaporisation of water = 2.260 kJ/g

2. To Vaporise the water at 100 ºC = 136.4 g x 2.260 = 308.3 kJ

Specific heat of steam = 0.00203 kJ/g/ºC

3. To heat the steam from 100 ºC to 125.4 ºC.

= 136.4 x 0.00203 x 25.4 = 7.033 kJ

Total heat absorbed = 44.98 + 308.3 + 7.033 = 360.3 kJ

Answer 2

We solve this problem in three steps:
(i) first, we have to heat the water to 100 C,
(ii) then we vaporize the water to get a steam at 100 C and,
(iii) finally, we heat the steam to 125.4 C.

The quantities of heat corresponding to each of these steps are:
Q(i) = m x c(water) x ΔT(water) = 0.1364 kg x 4190 J/kgK x (100 ─ 21.1)K = 45.093 kJ
Q(ii) = m x L(v) = 0.1364 kg x 2.26 MJ/kg = 308.264 kJ
Q(iii) = m x c(steam) x ΔT(steam) = 0.1364 kg x 2.030 kJ/kgK x (125.4 ─ 100)K = 7.033 kJ
The net heat that was absorbed in this process is then
Q(i) + Q(ii) + Q(iii) = 320 kJ