# Easy physics problem?!?!?

Boiled water that weighs 3kg is in an aluminium bowl that weighs 300g and the water has cooled to 17°C. How much heat was let free?

~I translated the problem, so it might not be written right, but it's correct.

I think I need to use the thermodynamic formula: m1c1(tf-t1)=m2c2(t2-tf)

Thanks from advance! :)

### 1 Answer

- ArjanLv 46 years agoFavorite Answer
You're looking in the right direction with those formulas.

Q=m*c*dT

with Q energy of the heat in Joule [J]

with m the mass of the object [kg]

with c the specific heat capacity in Joule per kilogramKelvin [J/(kg*K)]

with dT the temperature difference in Kelvin [K]

The temperature starts at 373 Kelvin (cause it's boiling) and it ends at 290 Kelvin (17°C)

so dT= 373-290=83

c is different for each material so you need to look this up:

c for water=4.2 J/(kg*K)

c for aluminium= 910 J/(kg*K)

Heat for the bowl:

Q=0.3*910*83= 22659 J

Heat for the water:

Q=3*4.2*83=1045,8 J

Total heat: 22659+1045,8=23704,8 J