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?
1 answer:
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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