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600g of water is mixed with 20g of steam at 100 deg Celcius. What is the final temperature if the original temperature of the water is 25 deg Celcius

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    The 20 g of steam will transfer 20g*540 Cal/g of heat to the liquid water while condensing. That makes 10,800 Cal, which is enough to raise the liquid water temperature by 18 C, to 38 C. As a final step, calculate the equilbrium temp T when mixing 600g liquid H2O at 38 C with 20 g of liquid H2O at at 100 C.

    600(T - 38) = 20(100 - T)
    Solve for T

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