Question

How many calories are needed to raise the temperature of liquid water from 25 0 C to steam at 121 0 C?

Answer 1

For water

specfic heat (liquid)= 4.184 J/g·°C

specific heat (gas) = 2.02 J/g·°C

ΔH° vaporization =40.7 kJ/mol

boiling point 100.0°C.

let mass of water be 'x' gram

we can work the problem in 3 steps:

1) For in Change temperature of x g water from 25 C to 100 C:
q = m c DT = (x) X 4.184 X 75 =313.8x J =0.3138 x kJ

2) Heat to vaporize x g water:
x g / 18 g/mol =x/18mol X 40.7 kJ/mol = 2.261 x kJ

3) Heat to raise temperature of steam: from 100 C to 121 C
q = m c DT = x X 2.02 X 21 C = 42.42 x J =0.0424 x KJ

therefore , total heat required = (0.3138x +2.261x +0.0424 x) KJ =2.6172x KJ

now as , 1 calorie = 4.186 KJ

hence 2.6172 x KJ =0.6252x calorie

0.6252x calorie are required for the process.( remember x is the mass of water in grams )