Question

A 7kg glass bowl contains 16kg of punch at 25^oC. Two and a half kilograms of ice is added to the punch. The ice has an initial temperature of -20^oC. Assume no transfer of heat to the external environment when equilibrium is reached, all the ice has melted and the final temperature of the mixture is above 0^oC. Determine this temperature?

C_glass = 840 J.kg^-1.^oC

C_ice = 2000 J.kg^-1.^oC

Latent heat of fusion of water, L_w = 335,000 J/kg

Answer 1

Yes specific heat value of punch will be same as water's

Heat gained = Heat lost

Heat gained = Q1 + Q2 + Q3

Q1 = ice heat to -20 C to 0 C = mi*Ci*dT1

Q2 = ice melts at 0 C = mi*Lf

Q3 = water heats from 0 C to Teq C = mi*Cw*dT2

Heat lost = Q4 + Q5

Q4 = heat lost by punch from 25 C to Teq C = mp*Cw*dT3

Q5 = heat lost by bowl from 25 C to Teq C = mb*Cb*dT3

So,

mi*(Ci*dT1 + Lf + Cw*dT2) = mp*Cw*dT3 + mb*Cb*T3

Using the given values

2.5*(2010*20 + 3.34*10^5 + 4186*(T - 0)) = 16*4186*(25 - T) + 7*840*(25 - T)

Solving this equation

T = 10.63 C