A 5.00 g sample of water vapor, initially at 155°C is cooled at atmospheric pressure, producing ice at - 55°C. Calculate the amount of heat energy lost by the water sample in this process, in kJ. Use the following dа ta: specific heat capacity of ice is 2.09 J/g*K; specific heat capacity of liquid water is 4.18 J/g*K; specific heat capacity of water vapor is 1.84 J/g*K; heat of fusion of ice is 336 J/g; heat of vaporization of water is 2260 J/g.

total energy lost = 16.1 KJ

Explanation:

Heat of vapor to water = 1.84j/gK is - because have a loss of energy

Heat lost during cooling from 155 to 100 = 1.84x 5 x (155 + 273 - 100+273) = 506J

latent heat of steam = 2260J/g x 5g = 11300J

Heat lost during cooling from 100 to 0 = 4.18 x5 x (100 - 0) = 2090J

Heat due to fusion is = 336J/g x 5g = 1680J

heat to cool ice of 0 to - 55 = 2.09J/gK x 5g x (0 - (-55)) = 574.75J

the sum of all H = H of reaction

506J + 11300J + 2090J + 1680J + 574.75J = 16150.75J = 16.1KJ

total energy lost = 16.1 KJ