Chemistry - Thermodynamics

A piece of metal of mass 12 g at 104◦C is
placed in a calorimeter containing 45.4 g of
water at 22◦C. The final temperature of the
mixture is 71.5
◦C. What is the specific heat
capacity of the metal? Assume that there is
no energy lost to the surroundings.
Answer in units of J
g ·
◦ C

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  1. The sum of the heats gained is zero.

    heatgainedmetal+heatgained water=0
    12*Cm*(Tf-Tim)+45.4*Cw*(Tf-Tiw)=0

    Tf=71.5 Tim=104 Tiw=22
    solve for cm

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  2. I plug in the numbers and solve using a calculator, this is how i did it.

    12(71.5-104)+45(71.5-22)=1857.3 J/g C

    I still get the answer wrong.

    Any errors i am making?

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  3. Sorry for posting twice :(

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  4. of course you get the wrong answer. What happened to cm,cw in my equation. You are looking for cm. Do some algebra.

    12*Cm*(Tf-Tim)+45.4*Cw*(Tf-Tiw)=0

    Tf=71.5 Tim=104 Tiw=22
    solve for cm

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    2. 👎 0

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