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    Physics_Phailure's Avatar
    Physics_Phailure Posts: 1, Reputation: 1
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    #1

    Apr 3, 2007, 06:25 PM
    Heat Transfer - Physics
    a .500kg sample of water is at 15 degrees C in a calorimeter. A .0400kg mass of zinc at 115 degrees C is placed in the water. What is the final temp of the system if specific heat of zinc is 388J/kg/C and the specific heat of water is 4184J/kg/C?

    This was the exact question goven to me I think it has something to do with the formula Q=mcDeltaT and energy has to always be at a constant since it can't be either destroyed or created, BUT don't take my word on that.

    ANY help wrong or hopefully otherwise would be greatly appreciated.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Apr 3, 2007, 11:29 PM
    Well, the water has to gain the same energy as the zinc loses, and the system has to come to an equal temperature.

    Using these 2 pieces of information, you should be able to form equations and solve them.
    NITESH KUMAR SAHA's Avatar
    NITESH KUMAR SAHA Posts: 4, Reputation: 1
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    #3

    Apr 5, 2007, 04:08 AM
    Quote Originally Posted by Physics_Phailure
    a .500kg sample of water is at 15 degrees C in a calorimeter. a .0400kg mass of zinc at 115 degrees C is placed in the water. what is the final temp of the system if specific heat of zinc is 388J/kg/C and the specific heat of water is 4184J/kg/C?

    This was the exact question goven to me i think it has something to do with the formula Q=mcDeltaT and energy has to always be at a constant since it can't be either destroyed or created, BUT dont take my word on that.

    ANY help wrong or hopefully otherwise would be greatly appreciated.
    m1*s1*[T-(273+15)] = m2*s2*[273+115-T]

    (m1*s1+m2*s2)T = m2*s2*388 +m1*s1*288

    T= {(.04*388*388)+(.5*4184*288)}/(.04*388 +.5*288)

    = X kelvin

    =X-273 deg c
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