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change in entropy of the ice/water

A one kilogram block of ice melts at 0°C and slowly warms up to the room temperature 30° C.

(a) What is the change in entropy of the ice/water?

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/1 kg

detalS_water = Q/T = (540000)(4.186)/303 = 7460.20 J/K

Did I do this part correctly?

(b) What is the change in entropy of the room?

(c) Is this process reversible or irreversible? Why?

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Solution Preview

Please see the attached file.

one kilogram block of ice melts at 0° C and slowly warms up to the room temperature 30° C.

(a) What is the change in entropy of the ice/water?

Here's what I did:

Lice = Q/m = 80 cal/g = 80000 cal/1 kg

ΔSice = Q/T = (80000)(4.186)/273 = 334880 J/273 K = 1226.67 J/K
This is correct for the entropy of the ice cube as it melts into water at 0°C.

Lwater = Q/m = 540 ...

Solution Summary

It finds the change in entropy when the ice melts and warms up to room temperature. It also finds the change in entropy of the room. The solution is detailed and was rated '5/5' by the student who posted the questions.

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