Question

A large punch bowl holds 3.50 kg of lemonade (which is essentially water) at 23.0 ∘C. A 5.20×10−2-kg ice cube at -10.0 ∘C is placed in the lemonade. What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings (in degrees C). What is the amount of ice (if any) remaining? (in kg)

Answer 1

specific heat of water is 4.186 kJ/kgC

specific heat of ice is 2.06 kJ/kgC

heat of fusion of ice is 334 kJ/kg

Energy given up by the ice in melting and warming up to the final temp, T, equals that of the lemonade cooling off from 23º to T, assuming all of the ice has melted.

E1 = 2.06×0.052×10 + 334×0.052 + 4.186×0.052×T = 18.44 + 0.218T

E2 = 4.186×3.50×(20–T) = 293 – 14.65T

Equating both:

18.44 + 0.218T = 293 – 14.65T

T = 18.5º