Temperature change boiling water

[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?

 

[Astronuc]

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

 

[lilkrazyrae]

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

 

[Astronuc]

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.

 

[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?

 

[Astronuc]

The formula is correct, but check the specific heat of copper.

I found c (copper) = 0.385 or 0.386 KJ/kg-K

http://en.wikipedia.org/wiki/Specific_heat_capacity#Table_of_specific_heat_capacities

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/sphtt.html#c1

 

[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!