4.C.13

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

4.C.13

How do you break this problem down to solve for the ice melting and the water heating. The solution manual is not making sense?

An ice cube of mass 50.0 g at 0.0 8C is added to a glass containing 400.0 g of water at 45.0 8C. What is the final tempera- ture of the system (see Tables 4A.2 and 4C.1)? Assume that no heat is lost to the surroundings.

Jessica Booth 2F
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

Re: 4.C.13

Because energy isn't lost you know that q1+q2=0 which means that q1=-q2. q1 is the heat required to first melt the ice and then increase its temperature to the final temp. So $q_{1}=n\Delta H_{fus} +mC\Delta T$. Also, q2 is the heat lost by the water when it is cooled down. So, $q_{2}=mC\Delta T$. Remember that the final temperature is the same for both the water and the ice so then you can solve for the final temp.