Question

You place a 7.35 g piece of copper at 113.6 °C into a styrofoam coffee cup...

You place a 7.35 g piece of copper at 113.6 °C into a styrofoam coffee cup with 60.3 mL of water at 18.4 °C. What is the final temperature of the system?

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Answer #1

Let us denote water by symbol 1 and Copper by symbol 2
since density of water is 1 g/mL, mass will be same as volume
m1 = 60.3 g
T1 = 18.4 oC
C1 = 4.184 J/goC
m2 = 7.35 g
T2 = 113.6 oC
C2 = 0.385 J/goC
T = to be calculated

Let the final temperature be T oC
use:
heat lost by 2 = heat gained by 1
m2*C2*(T2-T) = m1*C1*(T-T1)
7.35*0.385*(113.6-T) = 60.3*4.184*(T-18.4)
2.8297*(113.6-T) = 252.2952*(T-18.4)
321.4596 - 2.8297*T = 252.2952*T - 4642.2317
T= 19.5 oC

Answer: 19.5 oC

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