## First Law and Calculating Final Temp (question 8.21)

MSkye Goldwater 2K
Posts: 19
Joined: Fri Sep 29, 2017 7:05 am

### First Law and Calculating Final Temp (question 8.21)

Question 8.21 states, "A piece of copper of mass 20.0 g at 100.0 degrees C is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 degrees C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings."

The solutions manual starts with the statement "heat lost by metal = -heat gained by water"

Could I also say that q(metal) = -q(water) ?

Nisarg Shah 1C
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

### Re: First Law and Calculating Final Temp (question 8.21)

Yes, I believe the notation "heat lost by metal = -heat gained by water" is equivalent to q(metal) = -q(water).

Mia Navarro 1D
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

### Re: First Law and Calculating Final Temp (question 8.21)

To elaborate on why this is true, the energy released by the metal cannot escape the system, therefore it must transfer to the water. The negative sign serves to show that the energy gained by water is lossed by the metal and needs a negative sign to show that inverse relationship.

Diane Bui 2J
Posts: 61
Joined: Sat Jul 22, 2017 3:00 am

### Re: First Law and Calculating Final Temp (question 8.21)

Also to add on, I worked on this question with a TA and we started off with the general interpretation that -q(copper)=q(water) because heat is lost by the metal and gained by the water. We also established that the Tfinal for both copper and water would be the same.