Question

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 161.4 °C. What is the specific heat (c) of the substance?

Answer 1

Heat absorbed by the substance = 2717 J

Initial temperature of the substance (T_i) = 21.8 ^oC

Final temperature of the substance (T_f) = 161.4 ^oC

Temperature difference, $\Delta T$ = (T_f - T_i) = (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