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Q = m*c*\delta t\, where Q = heat flow, m = mass, c = specific heat capacity, and \delta t\ = change in temperature. 88.2Cal = 13.4g*c*153 degrees C. Simple algebra yields c = 0.043 Cal*m^-1*K^-1.

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14y ago
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14y ago

Since e=s*m*dt:

18.6=s*12*10.0

s=18.6/12/10.0 calories/(gram*degree Celsius)

s=0.155 calories/(gram*degree Celsius)

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Q: If it takes 18.6 calories of heat to raise the temperature of 12.0 g of a substance by 10.0 degrees Celsius what is the specific heat of the substance?
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