Answer:
[tex]0.985\ \text{J/g}^{\circ}\text{C}[/tex]
Explanation:
m = Mass of sample = 33.05 g
[tex]\Delta T[/tex] = Change in temperature = [tex]107.9-28.5=79.5^{\circ}\text{C}[/tex]
Q = Heat absorbed = 2589 J
c = Specific heat of substance
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow c=\dfrac{Q}{m\Delta T}\\\Rightarrow c=\dfrac{2589}{33.05\times 79.5}\\\Rightarrow c=0.985\ \text{J/g}^{\circ}\text{C}[/tex]
The specific heat of the substance is [tex]0.985\ \text{J/g}^{\circ}\text{C}[/tex].
