## 8.41

Alexandra Carpenter 1G
Posts: 32
Joined: Sat Jul 22, 2017 3:00 am

### 8.41

The question states, "A 50.0-g ice cube at 0.0 C is added to a glass containing 400.0 g of water at 45.0 C. What is the nal temperature of the system (see Tables 8.2 and 8.3)? Assume that no heat is lost to the surroundings."
The heat of the ice cube is 1.67 x 10^4 J + (209 J/degrees C)(Tfinal) and the heat of the water is (1.67 x 10^3 J/degrees C)(Tfinal)
The solution manual sets them equal to each other and shows the following: -(1.67 x 10^3 J/degrees C)Tfinal + 7.5x10^4 J = 1.67x10^4 J +(209 J/degrees C)(Tfinal)

Where did the +7.5x10^4 J come from??? Sorry this is so long.

Lisa Tang 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 2 times

### Re: 8.41

The 7.5 x 104 comes from the calculation of 400 grams water x (4.184 J/C/g) x (-45). From that I got -7.5 x 104, but because water is losing heat, you make qwater negative, turning the negative value positive (negative x negative= positive). Hope that helps.