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Chemistry 3 Online
OpenStudy (anonymous):

A 56.24 g sample of a substance is initially at 23.2 °C. After absorbing 2769 J of heat, the temperature of the substance is 105.5 °C. What is the specific heat (c) of the substance?

OpenStudy (aaronq):

use: \(q=m*C_p*\Delta T\) remember that \(\Delta T =(T_f-T_i)\)

OpenStudy (anonymous):

got it thank you!

OpenStudy (aaronq):

no problem

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