A 48-g ice cube floats in 220 g of water in a 100-g copper cup; all are at a tem

Question

A 48-g ice cube floats in 220 g of water in a 100-g copper cup; all are at a temperature of 0

A 48-g ice cube floats in 220 g of water in a 100-g copper cup; all are at a temperature of 0 degree C. A piece of lead at 94 degree C is dropped into the cup, and the final equilibrium temperature is 12 degree C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33 Times 105 J/kg and 4,186 J/kg ·degree C, respectively. The specific heat of lead and copper are 128 and 387 J/kg · degree C, respectively

Explanation / Answer

FInal equilibrium temperature = 12c

Heat energy released by lead = m*128*(94-12)= 10689*m J

Ice is converted to water.So energy absorbed by ice = 0.048*3.35*10^5 = 1680 J

energy absorbed by water lead and ice to raise to 12c is

(0.048+0.22)*4186*(12)+0.1*387*(12) = 94235.2 + 3250.8 = 13926 J

By conservation of enegy ewuating the terms gives

10689*m = 13926 + 1680

m = 1.46kg

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