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1 December, 11:18

A 40.27 g sample of a substance is initially at 24.8 °c. after absorbing 2099 j of heat, the temperature of the substance is 148.5 °c. what is the specific heat (c) of the substance?

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  1. 1 December, 11:27
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    Q = mCΔT

    Q is heat in joules, m is mass, C is specific heat, and delta T is change in temp

    2099 J = (40.27g) (C) (148.5 - 24.8) =.421 J / gram K
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