A 100 g copper calorimeter cup contains 235 cm3 of water and 45 g of ice, all in

Question

A 100 g copper calorimeter cup contains 235 cm3 of water and 45 g of ice, all in thermal equilibrium. A 240 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. A)What is the mass of water before the hot iron is put into the cup? Express your answer using three significant figures?? B)How much energy is required to melt all of the ice? (Signs matter!) C)How much energy is required to raise the temperature of the melted ice and water and copper cup to 30 C? (Signs matter!) D)How much energy leaves the iron in the form of heat? (Signs Matter!) E)What was the initial temperature of the iron?part A is 235 part B is 15.1 and part E is 508c idk C or D

Explanation / Answer

a)

Tf = final equilibrium temperature = 30 C

mc = mass of calorimeter = 100 g = 0.1 kg

mw = mass of water = density x volume = 1000 (235 x 10-6) = 0.235 kg

mi = mass of ice = 0.045 kg

mir = mass of iron = 0.24 kg

c)

heat required = (mw + mi) cw (30) = (0.235 + 0.045) (4186) (30) = 35162.4 J

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