# Latent heat

1. Jul 31, 2009

### tag16

1. The problem statement, all variables and given/known data
You are given 2.5 102 g of coffee (same specific heat as water) at 70.0°C. In order to cool this to 60.0°C, how much ice (at 0.0°C) must be added? Neglect heat content of the cup and heat exchanges with the surroundings.

2. Relevant equations
Q=(m ice)(c ice)(delta T ice+ m ice)(Lf)
Q=(m coffee)(c coffee)(delta T coffee)

3. The attempt at a solution
If these are the right formulas I still have the problem that there are 2 variables, since I don't know Q or m ice.

2. Jul 31, 2009

### mgb_phys

The Q's in both equations are the same, energy lost by coffee = energy gained by ice.
so you simply set the two equations equal to eahc other

3. Jul 31, 2009

### tag16

do you mean (m ice)(c ice)(delta T ice+ m ice)(Lf)=(m coffee)(c coffee)(delta T coffee)?

4. Jul 31, 2009

### mgb_phys

Almost, remember ther is energy to melt the ice (which doesn't depend on temperature) and then the energy to heat the resulting water to the final temperature.