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OpenStudy (anonymous):
A 34.54 g sample of a substance is initially at 23.2 °C. After absorbing 2873 J of heat, the temperature of the substance is 188.4 °C. What is the specific heat (c) of the substance?
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OpenStudy (cuanchi):
q= m x C x deltaT q=2873J m=34.54g deltaT= 188.4-23.2 isolate C, replace the variables in the formula and calculate C value
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