Question

You cool a 120.0 g slug of red-hot iron (temperature 745 ∘C) by dropping it into an insulated cup of negligible mass containing 75.0 g of water at 20.0 ∘C. Assume no heat exchange with the surroundings.

a) What is the final temperature of the water?

T final = ? degrees C

b) What is the final mass of the iron and the remaining water?

m final = ? g

Answer 1

Solution The amount of heat required to cool iron is = m,Ci (Tf-T, (0.120kg ) (450J/kgoC) (745°C-100°c) 34830J The amount of heat required to heat the water to boiling point is, AT (0.075 kg) (4179 J/kg°C) (100°C 20°c) - 25074J The remaining heat is O' = 34830-25074 = 9756 J Hence, the amount of heat required to evaporate water, since the final temperature of the water must be 100°C

The amount of mass evapomte is, Q=mL m=- 9756J 2264.76x103 J/kg =4.308×10-3 kg The total remaining mass of the iron and water is, = 120g +75 g-4.308 g = 1907 kg Hence, the final mass of the iron and the remaining water is 190.7g