Question

A large punch bowl holds 3.25 kg of lemonade (which is essentially water) at 20.0° C. A 1.90-kg ice cube at −10.2°C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.

Answer 1

heat lost by lemonande in reaching 0c , Q1 = Mlemonande*Sw*dt = 3.25*4190*(20-0) = 272350 J

heat gained by ice in melting , Q2 = Mice*Sice*dt + Mice*LIce = (1.9*2100*10.2)+(1.9*334*10^3) = 675298 J

Q1 < Q2

all the ice will not melt

ice and lemonand are in equilibrium at o0c

final temperature = 0oC

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amount of ice remaining = m

heat lost lemonande = heat gained by melted ice

Mlemonande*Sw*dt1 = Mice*Sice*dt + (Mice-m)*Lf

3.25*4190*20 = 1.9*2100*10.2 + (1.9 - m)*334*10^3

m = 1.2 kg

DONE please check the answer. any doubts feel free to ask