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46.9 g of a substance at 90.8°C is added to 51.8 g of water at 24.8°C. The final temperature of the water and substance is 40.5°C. The specific heat of water is 4.184 J/g·°C. What is the specific heat of the substance?

I think the equation that I need to use for this is C=Q/m * delta T. But I'm not sure

  • Chemistry - ,

    I think you are right but I like this way of doing it better.
    loss of heat from substance + heat gained by water = 0
    [mass substance x specific heat substance x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial) = 0 Solve for specific heat substance.

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