# Temperature change boiling water

1. Dec 11, 2005

### lilkrazyrae

An iron cylinder is heated to 100.degrees C in boiling water ant then transferred to 145.0 grams of water in a 70.0 gram copper calorimeter at 20.0 degrees C. If the final temperature is 25.2 degrees C find the mass of iron used.

Would I use mL(Boiling water)+mcT(water)+mcT(coper calorimeter) +mcT (iron)???

And what would be negative?

2. Dec 11, 2005

### Staff: Mentor

The iron will lose heat, so its temperature must decrease from 100°C to Tf = 25.2°C.

The 145 grams of water and the 70 g of copper in the calorimeter will experience a temperature increase from 20°C to Tf = 25.2°C.

3. Dec 11, 2005

### lilkrazyrae

Ok so is that the right equation because it keeps giving me a negative

4. Dec 11, 2005

### Staff: Mentor

Try, mc$\Delta{T}$(water)+mc$\Delta{T}$(copper calorimeter) +mc$\Delta{T}$ (iron) = 0.

The boiling water has nothing to do with the heat transfer. The only significant factor as far as boiling water is concerned is that the iron is heated to 100°C.

5. Dec 11, 2005

### lilkrazyrae

Ok so I've got (.1450)(4.186)(5.2) +(.070)(.85)(5.2) +(m)(.44)(-74.8)Which gives me 8.8194=2.912m and m=.255kg
So the iron cylinder is going to have more mass than the water and the calorimeter????

6. Dec 11, 2005

### Staff: Mentor

7. Dec 11, 2005

### lilkrazyrae

Oops! so (.1450)(4.186)(5.2) +(.070)(.385)(5.2) +(m)(.44)(-74.8) which gives me 3.296384=32.912m which gives the mass to be .100kg this makes more sense thanks!!