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A 50.0g piece of iron an 152°C is dropped into 20.0g H2O(l) at 89°C in an open, thermally insulated container. How much water would you expect to vaporize, assuming no water splashes out? The specific heats of iron and water are 0.45 and 4.21 respectively, and ÄHvap= 40.7 kJ mol^-1 H2O.
The iron will lose this much energy upon cooling to 100C:
.45 J*g^-1*K^-1 * 50g * 52K = 1170 J
The water will gain this much energy when heating to 100C
4.21 J*g^-1*K^-1 * 20g * 11K = 926.2 J
The difference in those energies will be used to vaporize water:
1170 J - 926.2 J = 243.8 J
To vaporize water, you need:
(40700 J/mol) / (18 g/mol) = 2261 J/g
243.8 J / (2261 J/g) = 0.1 g of water will vaporize