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90 g of water at 25C mixed with 50g of another substance at 100C result in 45C what is specific heat of that substance

  • physics -

    (M*C*delta T)of water + (M*C*delta T) of substance = 0
    90*1.0*20 + 50*C*(-55) = 0
    delta T (temperature change) of water is 20 C.
    delta T of unknown substance = -55 C

    Solve for the unknown specific heat C.

    C = 90*20/(50*55) = 0.654 cal/(g*C)

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