A 162 g piece of metal C 0371 Jg oC initially at 137 oC is

A 16.2 g piece of metal (C = 0.371 J/g oC) initially at 13.7 oC is placed in 40.5 g of water initially at 85.7 oC in an insulated container. Calculate the final temperature when the system comes to equilibrium. (C for water is 4..184 J/g oC.)

Solution

m(metal) = 16.3 g
T(metal) = 13.7 oC
C(metal) = 0.371 J/goC
m(water) = 40.5 g
T(water) = 85.7 oC
C(water) = 4.184 J/goC
T = 25.6 oC
We will be using heat conservation equation

Let the final temperature be T oC
use:
heat lost by water = heat gained by metal
m(water)*C(water)*(T(water)-T) = m(metal)*C(metal)*(T-T(metal))
40.5*4.184*(85.7-T) = 16.3*0.371*(T-13.7)
169.452*(85.7-T) = 6.0473*(T-13.7)
14522.0364 - 169.452*T = 6.0473*T - 82.848
T= 83.2 oC
Answer: 83.2 oC