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Question

A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the calorimeter is 31.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

Answer 1

You can use the formula q = mC∆T

The heat lost by the hot metal MUST be equal to the heat gained by the cooler water. This is in keeping with the law of conservation of energy.

heat lost by metal = q = mC∆T

q = heat

m = mass = 76.0 g

C = specific heat of the metal = ?

∆T = change in temperature = 96.0º - 31.0º = 65º

heat lost by metal = (76.0g)(C)(65º)

heat gained by water = q = mC∆T

q= heat

m = mass = 120.0 g

C = specific heat of water = 4.184 J/gº)

∆T = change in temperature = 31.0º - 24.5º = 6.5º

heat gained by water = (120.0 g)(4.184 J/gº)(6.5º) = 3264 J

Setting heat lost by metal equal to heat gained by water, we have ...

(76.0g)(C)(65º) = 3264 J

Solving for C (the specific heat of the metal, we have ...

C = 3264 J / (76.0 g)(65º)

C = 0.661 J/gº