Question

A 100 g copper calorimeter cup contains 240 cm3 of water and 25 g of ice, all in thermal equilibrium. A 230 g piece of iron is heated to a high temperature and dropped into the water (assume no water escapes the calorimeter cup). At thermal equilibrium the temperature of the entire mixture is 30 C.

(Questions posted below.)

Answer 1

a) What is the mass of water before the hot iron is put into the cup?
Express your answer using three significant figures in grams.

b) How much energy is required to melt all of the ice? Signs matter, answer in kJ.

c) How much energy is required to raise the temperature of the melted ice and water and copper cup to 30 C? Signs matter, answer in kJ.

d) How much energy leaves the iron in the form of heat? Signs matter, answer in kJ.

e) What was the initial temperature of the iron? Answer in C.

Answer 2

Heat gained by the calorimeter and the water and ice will equal the heat lost by the hot iron, all coming to a final temperature of 30oC.

Calorimeter gain will be (mass Cu)(specific heat Cu)(30oC) + (heat of fusion/melting of water)(mass of ice) + (mass of water and melted ice)specific heat of water)(30oC) = heat lost by iron, (mass of iron)(specific heat of iron)(T - 30)

Answer 3

a) Since the system is in thermal equilibrium, the water will not freeze and the ice will not melt so... we can use this formula