# Calorimetry with ice and water

1. Apr 12, 2010

### aal0315

1. The problem statement, all variables and given/known data
A block of ice at 0C is added to a 150g aluminum calorimeter cup which holds 210g of water at 12C. If all but 2.0 g of ice melt, what was the original mass of the block of ice

2. Relevant equations

heat gained by ice = heat lost by water + heat lost by calorimeter

3. The attempt at a solution
so i figure that its:
(m(i)+2.0x10^-3kg)*c(i)*deltaT(i) = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a), where m is mass, c is the specific heat and T is temperature
but what i dont get is isnt the deltaT(i) going to be zero since the ice starts off zero and the water/ice mixture temperature zero? so then this equation doesnt work because solving for m(i), you cant divide a number by zero?

2. Apr 12, 2010

### AtticusFinch

You need to use the latent heat of fusion to find out how much heat it takes to melt ice.

3. Apr 12, 2010

### aal0315

so instead of m(i)*c(i)*deltaT(i) in the above equation, I use m(i)*L, where L is 333 kJ/kg?

4. Apr 13, 2010

### AtticusFinch

Yeah, it's a phase change.

5. Apr 13, 2010

### aal0315

so i did what you said and this is what i got
(m(i)+2.0x10^-3kg)L = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a)
(m(i)+2.0x10^-3)(333) = (210x10^-3*4186*(12-0)) + (150x10^-3*900*(12-0))
m(i) = 36.54kg
this doesnt make sense does it??

6. Apr 13, 2010

### AtticusFinch

Wait, shouldn't your specific heats stay at the same numbers?

4.186 J/goC = 4.186 kJ/kgoC

7. Apr 14, 2010

### aal0315

what do you mean? little confused ... you said to use the m(i)*L for the left side of the equation and then for the right side, don't i use the specific heats for water and aluminum?

8. Apr 14, 2010

### AtticusFinch

You do, but you messed up the numbers. You used 4186 J/kgoC instead of 4.186 kJ/kgoC. So your answer comes out off by a factor of 1000.

Make sure you check that your units are consistent before you answer the question (you used 333 kJ/kg for latent heat so every value should involve kJ and/or kg).

Also, this term is incorrect. (m(i)+2.0x10^-3)(333) All the initial mass BUT 2 g melts so you should subtract.

9. Apr 14, 2010

### aal0315

alright .. i get it now .. thanks for the help!