calorimetry question

Question

A 138.7 g piece of copper (specific heat 0.38 J/g･°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g･°C).

Answer 1

Heat lost by the copper = heat gained by the water

Heat lost by copper = mC∆T = (138.7 g)(0.38 J/g/deg)(∆T) = 52.71 ∆T

Heat gained by water = mC∆T = (400.0 g)(4.18 J/g/deg)(22.2º - 20.7º) = 2508 J

Setting them equal, we have....

52.71 ∆T = 2508

∆T = 2508/52.71 = 47.6 degrees

This is the CHANGE in temperature for the copper. The final temperature of the copper is the same as the final temperature of the water (equilibrium), which is 22.2º C. Thus the initial temperature of the copper is...

47.6º + 22.2º = 69.8ºC = Initial temperature of the copper

Another way to do this is as follows:

(138.7 g)(0.38 J/g/deg)(Ti - 22.2) = (400.0 g)(4.18 J/g/deg)(22.2º - 20.7º)

52.71Ti - 1170 = 2508

52.71 Ti = 3678

Ti = 69.8 degrees