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A piece of lead with a mass of 29.3 g was heated to 97.85-degrees C and...

A piece of lead with a mass of 29.3 g was heated to 97.85-degrees C and then dropped into 16.0 g of water at 22.80-degrees C. The final temp was 26.61-degrees C. Calculate the specific heat capacity of lead from these data. (The specific heat capacity of liquid water is 4.184 J/g K).

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Answer #1

heat absorbed by water = heat lost by lead metal

so

heat absorbed by water (q) = m*S*DT

                                        = (16)*4.184*(26.61 -22.8)

                                        = 255.1 J

heat lost by lead metal = 255.1 J

255.1 J = m*S*DT

255.1 J = 29.3*S*(97.85 - 26.61)

S = 0.12 J/g K

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