Determine the specific heat of lead, knowing that lead weighing 100 g at a temperature of 100 degrees

| August 7, 2021 | education

Determine the specific heat of lead, knowing that lead weighing 100 g at a temperature of 100 degrees is immersed in an aluminum calorimeter weighing 40 g, containing 240 g of water at a temperature of 15 degrees, then a temperature of 16 degrees is set in the calorimeter.

Given: m1 (lead) = 100 g (0.1 kg); t1 (initial temperature of lead) = 100 ºС; m2 (calorimeter) = 40 g (0.04 kg); m3 (water) = 240 g (0.24 kg); t2 (initial temperature of the calorimeter, water) = 15 ºС; t3 (set temp.) = 16 ° C.

Constants: С2 (aluminum) = 920 J / (kg * ºС); C3 (water) = 4200 J / (kg * ºС).

Expression: C1 * m1 * (t1 – t3) = (C2 * m2 + C3 * m3) * (t3 – t2), whence C1 = (C2 * m2 + C3 * m3) * (t3 – t2) / (m1 * (t1 – t3)).

Calculation: C1 = (920 * 0.04 + 4200 * 0.24) * (16 – 15) / (0.1 * (100 – 16)) = 124.38 J / (kg * ºС).



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