Entropy change: Melt ice-water and heating water fr 0C-25C

100 g of ice is melted at 0 degree C, and the resulting water is warmed to room temperature (25 degree C). Calculate delta S (change in entropy) for this whole process and comment on the relative magnitude of the entropy changes for the two steps. Explain the sign of delta S.

Solution Preview

Basically what you have to consider here, is two changes:

1) State change Ice to Water (a latent heat change).

2) Heating of the water and the associated temperature change.

A state change where ice is melted i.e. a change from solid (ice) to liquid (water), you will need to determine the amount of heat required for this change.

Call this heat the Latent Heat Delta.Q1 which can be calculated by

Delta.Q1 = ML ,

where M is the mass of the ice in kg which in this case M = 0.1 kg, L is the specific latent heat of water which you will need to ...

Solution Summary

A step by step path to a solution to show how to determine the entropy changes by heating and melting 100g of water ice to water and heating the water from 0 to 25 C is shown. This is developed along the paths of first considering a latent heat change and determining the change in heat, thus the change in entropy of this change and the considering the heating of the water and the heat change and associated entropy change with this process.

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Here's what I did:
L_ice = Q/m = 80 cal/g = 80000 cal/1 kg
deltaS_ice = Q/T = (80000)(4.186)/273 = 334880 J/273 K = 1226.67 J/K
L_water = Q/m = 540 cal/1 g = 540000 cal

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