How much heat is involved in converting 456g of H2O(g) at 127 degree C and to H2O(l) at 40 degree C.

Question

Specific heat of H2O(g) = 1.90J/g degree C and H2O(l) = 4.18J/g degree C. ΔH fus: 334J/g. ΔHvap: 2.26kJ/g.

Answer 1

The first step is to cool the gas from 127° C to 100° C.  To find this, multiply the specific heat (1.90 J/g-°C) times the mass (456 g) and the change in temperature (27° C).  This gives:

 

1.90 * 456 * 27 = 23392.8 J

 

The second step is to convert the gas to a liquid.  To find this, multiply the heat of vaporization (2.26 kJ/g = 2260 J/g) times the mass (456 g).  This gives:

 

2260 * 456 = 1030560 J

 

Finally the liquid must be cooled from 100° C to 40° C, a change of 60° C.  This is done the same way as the first step, except that the specific heat for the liquid is different -- 4.18 J/g-° C.  So now we get:

 

4.18 * 456 * 60 = 114364.8 J

 

At every step, energy is released.

 

To find the total amount of energy released, add the amounts from each step, which comes to 1168317.6 J = 1168.3 kJ.

 

Since the question asked specifically about "heat," you might want to convert this to calories or kilocalories (though using Joules should be OK).  This can be done with the conversion constant of 0.2389 calories per Joule, which results in 279111 calories, or 279.111 kcal.