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An object at 20 degrees Celsius with heat capacity 100 calories/Celsius is placed in 40 grams of water at 100 degrees Celsius. After equilibrium is reached, the object is removed from the hot water and placed in 200 grams of water at 0 degrees Celsius. What is the final temperature of the object? Explain.

Thanks for the help!

  • Physics -

    You will have t do this problem in two steps.

    In step 1, determine the equilibrium temperature when the object is placed in the 100 C water. The heat lost by the water will equal the heat gained by the object. Write that equation and solve for the final temperature of that step, T1.

    (100 cal/C)*(T1-20)
    = (40 Cal/C)*(100 - T1)
    Solve for T1

    (100-T1)/T1-20) = 2.5
    100 - T1 = 2.5 T1 -50
    3.5 T1 = 150
    T1 = 42.8 C

    Once you have T1, write an analogous equation for the process of reaching thermal equilibrium in the second step. Then solve for the final equilbrium temperature, T2.

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