A worker pours 1.250 kg of molten lead at a temperature of 327.3°C into 0.5000 kg of water at a temperature of 75.00°C in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

Solution 105P Step 1: The mass of molten lead is 1250 kg. The mass of water is 0.5000 kg. The temperature of lead is 327.3 C . 0 The temperature of water is 75 C . 1 1 Specific heat of water is 4190 J kg K . 1 1 Specific heat of lead is 130 J kg K . The heat lost by the lead is equal to heat gained by water. So, Q lost mcT = 1250 × 130 × (327.3 T) = 162500 (327.3 T) Qgain= mcT = 0.500 × 4190 × (T 75)) = 2095 (T 75) So, 162500 (327.3 T) = 2095 (T 75) (2095 + 162500) T = 162500 × 327.3 2095 × 75 164595 T = 53029125 0 T = 322.17 or 595.32 K .