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A 31.29 g sample of a substance is initially at 21.8 °C. After absorbing 2717 J of heat, the temperature of the substance is

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Answer #1

Heat absorbed by the substance = 2717 J

Initial temperature of the substance (Ti) = 21.8 oC

Final temperature of the substance (Tf) = 161.4 oC

Temperature difference, \Delta T = (Tf - Ti) = (161.4 oC - 21.8 oC) = 139.6 oC

Mass of the substance (m) = 31.29 g

Let us say that the specific heat capacity of the substance = s

Now, heat energy absorbed = m x s x \Delta T                 

or, 2717 J = 31.29 g x s x 139.6 oC

or, s = 0.6220 J/g.oC

Hence, the specific heat of the substance = 0.6220 J/g.oC

                                           

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