Chemistry aktiv eq

Question

A 117.5 g piece of copper (specific heat 0.380 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 a final temperature of 22.2 °C. What is the initial temperature (in °C) of the copper? (The specific heat of water is 4.184 J/g･ °C).

Answer 1

Heat lost by copper must equal heat gained by water (conservation of energy)

heat lost by copper = q = mC∆T

m = mass = 117.5 g

C = specific heat = 0.380 J/gº

∆T = change in temperature = Ti - 22.2º where Ti is the initial temperature of the copper

q = (117.5 g)(0.380 J/gº)(Ti - 22.2)

heat gained by water = q = mC∆T

m = mass = 400.0 g

C = 4184 J/gº

∆T = change in temperature = 22.2 - 20.7 = 1.5º

q = (400.0 g)(4.184 J/gº)(1.5º) = 2510 J

Setting the two heats equal, we have...

2510 J = (117.5 g)(0.380 J/gº)(Ti - 22.2)

2510 = 44.65Ti - 991

44.65 Ti = 3501

Ti = 78.4ºC = initial temperature of the copper