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

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  • Arjan
    Lv 4
    6 years ago
    Favorite 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

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