5 g of water at 30oC and 5 g of ice at -20oC are mixed together in a calorimeter. Find the final temperature of the mixture. Assume water equivalent of calorimeter to be negligible, sp. heats of ice and water are 0.5 and 1 cal/gCo , and latent heat of ice is 80 cal/g.

Here ice will absorb heat while hot water will release it. So if T is the final temperature of the mixture, heat given by waterQ1=mcΔT=5×1×(30−T)And heat absorbed by iceQ2=5×(1/2)[0−(−20)]+5×80+5×1(T−0)So, by principle of calorimetry Q1=Q2,i.e.,      150−5T=450+5TT=−30∘Cwhich is impossible as a body cannot be cooled to a temperature below the temperature of cooling body. The physical reason for this discrepancy is the heat remaining after changing the temperature of ice from -20 to 0oC with some ice left unmelted and we are taking it for granted that heat is transferred from water at 0oC to ice at 0oC so that temperature of system drops below 0oC. However, as heat cannot flow from one body (water) to the other (ice) at same temperature (0oC), the temperature of system will not fall below 0oC.ALTERNATE METHOD :When water and ice are mixed  if the heat lost by water is insufficient to melt the ice then the resultant temperature is equal to 00 c