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A 20-g ice cube floats in 210 g of water in a 100-g copper cup; all...

A 20-g ice cube floats in 210 g of water in a 100-g copper cup; all are at a temperature of 0°C. A piece of lead at 92°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33 105 J/kg and 4,186 J/kg · °C, respectively. The specific heat of lead and copper are 128 and 387 J/kg · °C, respectively.)

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Answer #1

mice = mass of ice = 0.020 kg

mw = mass of water = 0.210 kg

mc = mass oc copper = 0.100 kg

Heat lost by lead = mlead x Clead (92 - 12) = m (128) (80) = 10240 m

Heat gained = mice L + mice Cwater (12 - 0) + mw Cw (12 - 0) + mc Cc (12 - 0)

Heat gained = 0.020 (3.33 x 105) + 0.020 (4186) (12) + 0.210 (4186) (12) + (0.1) (387) (12)

using conservation ogf energy

Heat lost = Heat gained

10240 m = 0.020 (3.33 x 105) + 0.020 (4186) (12) + 0.210 (4186) (12) + (0.1) (387) (12)

m = 1.824 kg

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