Answer
c = 1.380 J/g°C
Explanation
Given:
Mass of sample, m = 10 g
Initial temperature, T₁ = 10 °C
Final temperature, T₂ = 81 °C
Change in temperature, ΔT = T₂ - T₁ = 81 °C - 10 °C = 71 °C
Quantity of heat, Q = 980J
What to find:
The specific heat, c of the material.
Step-by-step solution:
The specific heat, c of the material can be calculated using the formula given below:
[tex]\begin{gathered} Q=mc\Delta T \\ \\ \Rightarrow c=\frac{Q}{m\Delta T} \end{gathered}[/tex]Plugging the values of the parameter, we have;
[tex]c=\frac{980J}{10g\times71°C}=\frac{980J}{710\text{ }g°C}=1.380\text{ }J\text{/}g°C[/tex]Therefore, the specific heat, c of the material is 1.380 J/g°C
