A 34-g ice cube floats in 240 g of water in a 100-g copper cup; all are at a tem

Question

A 34-g ice cube floats in 240 g of water in a 100-g copper cup; all are at a temperature of 0

A 34-g ice cube floats in 240 g of water in a 100-g copper cup; all are at a temperature of 0A degree C. A piece of lead at 86A degree C is dropped into the cup, and the final equilibrium temperature is 12A degree C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33 x 10^5 J/kg and 4,186 J/kg A? A degree C, respectively. The specific heat of lead and copper are 128 and 387 J/kg A ? A degree C, respectively.) kg

Explanation / Answer

Energy released by the lead = energy gained by the (ice+water+copper cup)

mCdT = energy for ice to melt + energy for mass of (melted ice +water) to reach 12 C +energy for cup to reach 12 C

m*128*(86-12) = M-ice*heat of fusion + (mass of melted ice + water)*Sp heat water*dT + m-copper*C*dT

9472*m = 0.034*3.33*10^5 + (0.034+0.240)*4186*(12-0) + 0.100*387*(12-0)

9472*m = 11322 + 13763.57 + 464.4 = 25549.97

m = 2.7 kg

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