# Homework Help: Calorimetry Problem

1. Jan 22, 2010

### sdoug041

1. The problem statement, all variables and given/known data

You cool a 90.0 g slug of red-hot iron (temperature 745 C) by dropping it into an insulated cup of negligible mass containing 85.0 g of water at 20.0 C. Assume no heat exchange with the surroundings.

What is the final temperature of the water?

2. Relevant equations

q = mc$$\Delta$$T and most likely q = mL

3. The attempt at a solution

I know this is a fairly straight forward question but I am not receiving the right value for the temperature of the water. I use the formula q(water) = -q(iron) and solve for T(2). I keep getting back 97C, but the correct answer (apparently) should be 100C because for the next part of the question you need to figure out how much water is left over from some of it converting to steam.

Here is some additional information: The specific heat capacity I'm using for Iron is 470 J/kgK and for water 4190 J/kgK

2. Jan 22, 2010

### fluidistic

I suggest you to use some intuition, or as you did, reading the next part of the question. It is clear that water will reach 100 °C and a portion of it will steam.
So writing down $$Q_{\text{water}}+Q_{\text{iron}}=0$$ is not appropriated, because you don't take in count the energy required to steam up the water.
You have to add a term in the equation, the one you wrote under the "Revelant equations".
Good luck!
Edit: You could also do some research on the forum. There are many, many similar problems in the database.

3. Jan 22, 2010

### sdoug041

But shouldn't I be yielding a value >100 if I use $$Q_{\text{water}}+Q_{\text{iron}}=0$$? You can't just know that it's going to be greater than 100 then add the additional term... Aren't you supposed to test that by using the equation to see if you get over 100? And then if you do you need to take in consideration the thermal energy used to produce the steam?

4. Jan 22, 2010

### Staff: Mentor

I got 97 deg C as well.

That is assuming specific heat of iron is 470 J/kgK and is not temperature dependent.

Edit: and I wouldn't worry about steam. As long as the answer is BELOW 100 deg C, we know we are on the safe side. Once the answer becomes 103 deg C, we know our equation for heat balance was wrong.

Edit 2: that's exactly what you wrote - and you are right about it.

5. Jan 22, 2010

### fluidistic

You're right. That's why I said I'd glad to see your derivation.
Oh well, my intuition is dead wrong, once more!

6. Jan 22, 2010

### sdoug041

Alright Well thanks guys I'm just going to assume my prof entered the wrong value for now.