If 12.0 gram of ice are dropped in 150. ml of water at 45 degrees celcius, what is the final temperature of the water?

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  1. Note the correct spelling of celsius.
    I assume the temperature of the ice is zero C.
    heat gained by ice + heat gained by melted ice + heat lost by 45 degree water = 0.
    [mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) = 0
    Solve for Tfinal. I get something like 36 degrees but that's an approximation.

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  2. Filling in the equation, here is what I have. Still not coming up with 36 degrees. Please help:

    [mass ice x heat fusion]
    12.0g x 334j/g

    [mass melted ice x specific heat water x (Tfinal-Tinitial)]
    12.0g x 4.18j/g x 45 (Delta T is 45 because temp of ice from 0 degrees to water at 45 degrees)

    [mass water x specific heat water x (Tfinal-Tinitial)
    150g x 4.18j/g x (Tf - 45)

    Also, how would I go about solving for Tfinal?

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  3. After some sleep, i was able to come up with the following. It's still not 36 degrees, but close. Can some check my calculation?

    [mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) =0
    12.0g x 334j/g = 4008j + 12.0g x 4.18j/gC x 45C =2257j + 150g x 4.18j/gC x (Tf – 45C) = 0
    -6265j = 150g x 4.18j/gC x (Tf – 45C)
    -6265 = 627C x (Tf – 45C)
    -9.99 C = Tf – 45C
    Tf = 35.01C

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  4. The middle term is the one that is not right. 12.0 g x 4.184 x (Tfinal-Tinitial). Tfinal is Tfinal no matter where you see it; therefore, Tfinal is what you solve for. You have one Tfinal in the middle term and another in the last term. As far as delta T goes for the middle terms, that is not 45 (45-0) but Tfinal-0). The melted ice starts at zero C and goes to Tfinal.

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  5. Makes perfect sense. Thank you.

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