Question

If 150 g of water at 30 deg C is mixed with 300 g of Ice at 0 deg C, the total change in entropy will be

Properties of Water:
Heat of Fusion hF334 J/g
Heat of VaporizationHV2264.76 J/g
Specific Heat of IceCice2.080 J/g/K
Specific Heat of WaterCwater4.1813 J/g/K
Specific Heat of VaporCvapor2.11 J/g/K

a.

18800 J/K

b.

298 J/K

c.

68.9 J/K

d.

3.53 J/K

Answer 1

Heat lost by water to while cooling down to 0 degC = 150*4.18*30 = 18810
Heat required to melt 300 gm ice = 300*334 = 100200
As heat from water if less than heat needed to melt the 300 gm ice, 18810 j of heat will be transferred form water to ice and final tempreture of the system will be Ice + water at 0 deg C.
Increase in entropy of ice = heat gained / tempreture
                                         = 18810/273 = 68.9 /K
decrease in entropy of water =

where m is mass of water and c is it's specifice heat.
Taking limits of integral from 30 degC(303K) to 0 degC(273 K), we get
150*4.18 * = 65.37 /K

Net increase in entropy = 68.9-65.37 = 3.53 /K