Assume that the heat capacity of the calorimeter equals the heat capacity of the water it contains. Calculate the specific heat capacity of the metal and use the result to identify the metal.

Question

A 44.0 g sample of an unknown metal at 100.0 °C is placed in a constant pressure calorimeter containing 80.0 g of water at 24.8 °C. The final temperature is 28.4 °C. Assume that the heat capacity of the calorimeter equals the heat capacity of the water it contains. Calculate the specific heat capacity of the metal and use the result to identify the metal.

Answer 1

Heat lost by hot metal must equal heat gained by water/calorimeter (conservation of energy).

heat gained by water = q = mC∆T

q = heat = ?

m = mass of water = 80.0 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature of water = 28.4º - 24.8º = 3.6º

q = (80.0 g)(4.184 J/gº)(3.6º) = 1205 J

heat lost by metal = q = 1205 J

q = mC∆T

q = heat = 1205 J

m = mass of metal = 44.0 g

C = specific heat of metal = ?

∆T = change in temperature of the metal = 100º - 28.4º = 71.6º

1205 J = (44.0 g)(C)(71.6º)

C = 1205 J / (44.0 g)(71.6º)

C = 0.382 J/gº

To use this to identify the metal, you should have been given a table of heat capacity values for different metals. See which one most closely aligns with 0.382 J/gº.