# Homework Help: Heat Capacity and Latent Heat question

1. Oct 14, 2016

### stratz

1. The problem statement, all variables and given/known data
A 75.0g ice cube at 0 C is placed in 825g of water at 25 C. Find final temp.

2. Relevant equations
Well none really, this should be simple plugging into to heat equations.

3. The attempt at a solution

Heat required to melt ice = (0.075kg)(3.33*10^5) = 24975 J

Heat required to melt ice + Heat required to raise temp of melted ice = heat released from dropping temp of water

24975 + (0.075kg)(4186J/kg C)(Tf) = (0.825)(4186J/kg C)(Tf - 25)

Rearrange the equation: Tf ends up being 35.4, which is impossible. What am I doing wrong?

2. Oct 14, 2016

### Staff: Mentor

You are expecting the final temperature to be lower than 25C, right? Take a look at your ΔT for the water.

3. Oct 14, 2016

### stratz

The Tf-25 right? Shouldn't it be that way because the delta T would equal Tf-Ti? Or is the issue that I am supposed to put a negative 1 coefficient in front of the whole sign for Qg = -Ql ? If I do that I get Tf as a negative number though.

4. Oct 14, 2016

### Staff: Mentor

If you write ΔT = (25 - Tf) and Tf is less than 25 (as you expect it to be) then the heat quantity you calculate will be a positive value. On the other side of the equation you're raising the temperature of the ice water from zero to Tf, and you wrote its ΔT as (Tf - 0). You are equating two positive values, which is fine.

Suppose you were to move the liquid water expression to the left hand side of the equals so that you sum the two heats. You expect that sum to be zero since no heat is being magically created or destroyed, it's just being moved around. If you move the RHS to the left then, to change its sign you can change the signs in its ΔT term, making it now (Tf - 25), and its corresponding heat will be negative, representing a heat loss.