Answer to Question #304093 in General Chemistry for shan

Question

Compare the cooling effect of drinking 250 ml of ice water with the cooling effect of sweating out 250mL of water

Answer 1

First you have to decide on the processes taking place - how many changes of state are there and how many ‘heating up’ processes.

1. Ice at 0 deg to water at 0 deg (change of state)
2. Water at 0 deg heated to water at 100 deg (heating up)
3. water at 100 deg to water vapour at 100 deg (change of state)

1 and 3 both involve the appropriate latent heat - latent heat of melting (aka latent heat of fusion) and latent heat of vaporisation.

2 involves the specific heat capacity of water

For 1: heat energy = mass * latent of heat of melting (ice)

For 3: similarly, heat energy = mass * latent heat of vaporisation (water)

For 2: heat energy = c * m * change in temperature (c is specific heat of water)

Now, go and look up the unknown numbers, do the (simple) arithmetic and add up the three heat energies you’ve calculated for your final answer.