Caliometry calculation

Question

A piece of metal weighing 48.045 g was heated to 100.0 °C and then put it into 200.0 mL of water (initially at 25.0 °C). The metal and water were allowed to come to an equilibrium temperature, determined to be 29.2 °C. Assuming no heat lost to the environment, calculate the specific heat of the metal.

Answer 1

heat lost by metal = heat gained by water

heat lost by metal = q = mC∆T where q = heat; m=mass; C=specific heat; ∆T=change in temperature

q = (48.045g)(C)(100º - 29.2º) = (48.045)(70.8)(C) = 3401C

heat gained by water = q = mC∆T

q = (200.0g)(4.184 J/g/deg)(4.2) = 3515 J (assuming a density of 1g/ml for water)

3401C = 3515

C = 1.03 J/g/degree = specific heat of the metal