Can someone help me with this chemistry 馃И problem?

A 59.4 g sample of metal at 90.81 oC is added to 56.38 g of water that is initially at 16.84 oC. The final temperature of both the water and the metal is 29.36 oC. The specific heat of water is 4.184 J/(goC). Calculate the specific heat of the metal.

1 Answer

  • 1 month ago
    Favorite Answer

    (4.184 J/g路掳C) x (56.38 g) x (29.36 - 16.84)掳C =

    2953.39188 J gained by the water (and lost by the metal)

    (2953.39188 J) / (59.4 g) / (90.81 - 29.36)掳C = 0.809 J/g路掳C

    [Although it is difficult to think of an actual metal with a specific heat like that.]

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