Question

A piece of copper metal is initially at 83.0°C. It is dropped into a coffee cup calorimeter containing 30.0 9 of water at a temperature of 10.0°c. After stirring, the final temperature of both copper and water is 25.0°c. Assuming no heat losses, and that the specific heat (capacity) of water is 4.18 J/(g.), what is the heat capacity of the copper in J/K?

Answer 1

The initial temperature of Copper, T1 = 83^oC = 83+ 273 = 356 K
Final temperature of Copper, T = 25^oC = 25+273 = 298 K
Initial temperature of Calorimeter, T2 = 10^oC = 10+273 = 283K
Mass of water in calorimeter, m = 30g
Specific heat of water, s = 4.18 J / (g.K)

Now,
heat lost by copper peice = heat gained by calorimeter
C*(T1 - T) = m*s*(T - T2)
where, C is heat capacity of copper in J/K
therefore, putting values
C * (356K - 298K) = 30g * 4.18 /(g.K) * (298K - 283K)
C*58K = 30g*4.18 J/(g.K) * 15K
C * 58 = 1881
C = 1881/58
C = 32.43 J/K
thus, heat capacity of copper is 20.22 J/K