A vessel which contains an ice-water mixture at 0C is in thermal contact with a

Question

A vessel which contains an ice-water mixture at 0C is in thermal contact with a water-steam mixture at 100 C.

A.During a 10 minute period, 35 grams of steam condenses into water. How much energy flowed out of the water-steam mixture? (Signs Matter!) kJ

B.Assuming that all of the energy flowing out of the water-steam mixture goes into the water-ice mixture, how many grams of ice melts?

C.What is the change in the entropy of the water-steam mixture at 100C? (Signs Matter!)

D. What is the change in the entropy of the water-ice mixture at 0C? (Signs Matter!) (Verify to yourself that the 2nd law of thermodynamics is satisfied)

Explanation / Answer

latent heat of condensation of steam L = 2260 kJ/kg

heat evolved = mass of steam x L = 35/1000 x 2260 = - 79.1 kJ

this much energy flowed out so there will be a negative sign

now

latent heat of fusion of ice = 334 kJ/kg

mass of ice melted x 334 kJ/kg = 79.1 kJ

mass of ice melted = 0.2368 kg

entropy change for steam

dS = Q/T = 79.1 / 373 = 212.06 J

entropy change for ice

dS = Q/T = 79.1 / 273 = 289.743 J

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