## Chapter 8 HMWK Problem 8.21 Clarification

$\Delta U=q+w$

Sunny Chera 1N
Posts: 20
Joined: Wed Sep 21, 2016 2:56 pm

### Chapter 8 HMWK Problem 8.21 Clarification

The question is stated as follows:

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

Problem: I know that the heat lost by the metal is equal to the heat gained by the water. Therefore, I said that -q=qH20. Although I got the same answer as in the solutions manual, it stated that the heat lost by the metal is equal to the negative heat gained by water (qmetal=-qH20). Is either interpretation considered valid, or is there a flaw with my reasoning? I've had the same issue with several other problems as well.

Aishwarya_Natarajan_2F
Posts: 11
Joined: Mon Jul 11, 2016 3:00 am

### Re: Chapter 8 HMWK Problem 8.21 Clarification

In a case of heat transfer between a system and it's surroundings, we assume qsystem +qsurrounding= 0, or in other words q sys = -q sur

In this case the problem is asking us to examine heat transfer between a piece of copper at a temp of 100 deg C (the system) and the water at 22.0 deg C (the surroundings).

You are correct in your setup of -qmetal = qH2O because when you plug values into this you get:

-(20.0 g)(0.38 J/g C)(Tfinal - 100 C) = (50.7 g)(4.184 J/g C)(Tfinal - 22)

Logically, we know that the final temperature will lie somewhere in the middle of 100 and 22 when this transfer occurs, which means, ignoring the negative sign, that the q sur, or water side of the equation, will be positive because it's temperature will rise, and the q sys, or copper side, will be negative because it's temperature falls. If you put the negative sign on the water side, then both sides will become negative and if you put it on the copper side, both sides will become positive. Mathematically, having the negative sign there is like multiplying one of the sides by (-1), and so in this case it actually doesn't matter what side the negative sign is on. This is why in this problem, either interpretation would be valid.