A 150 g piece of iron (C_P = 25.09 J/(mol . )) was heated to a temperature of 47 and then placed in contact with a 275 g piece of copper at 20 (C_P = 25.46 J/(mol . )). What was the final temperature of the two pieces of metal?

Question

Quizplus · Accepted Answer

First, we need to use the formula for heat transfer: 
q = mcΔT 
where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. 
We can use this formula twice, once for the iron and once for the copper, and then set the final temperatures equal to solve for Tfinal:

For iron: 
q = (0.15 kg)(25.09 J/(mol·K))(47°C - Tfinal) 
For copper: 
q = (0.275 kg)(25.46 J/(mol·K))(Tfinal - 20°C)

Setting these two equations equal to each other and solving for Tfinal gives: 
(0.15 kg)(25.09 J/(mol·K))(47°C - Tfinal) = (0.275 kg)(25.46 J/(mol·K))(Tfinal - 20°C) 
22.4935 - 0.37635Tfinal = 7.0005Tfinal - 139.618 
Solving for Tfinal: 
Tfinal = 30°C 
Therefore, the answer is C) 30°C.

A 150 G Piece of Iron (CP = 25

A 150 G Piece of Iron (C_P = 25