Chapter 15.2, Problem 21SSC

Interpretation Introduction

Interpretation:

Thespecific heat of an unknown substance when a $2.5g$ sample releases $12\mathrm{cal}$ with a temperature change from $25°C$ to $20°C$ is to be calculated.

Concept introduction:

The heat absorbed or released by a substance is dependent on three factors:

1. The mass of the substance.
2. The nature of the substance, its specific heat capacity.
3. The change in temperature of the substance.

The SI unit of heat is Joules (J), specific heat is J/(kg degree Celsius), and that of mass is kg.

Answer to Problem 21SSC

The mathematical formula to calculate the heat absorbed or released by a substance when the temperature changes is,

$Q=m\times c\times \Delta T$

Where,

$Q$ is the heat released or absorbed,

$m$ is the mass of the substance,

$c$ is the specific heat of the substance,

$\Delta T$ is the difference between the initial and the final temperatures.

Given,

$\begin{array}{l}{T}_{initial}=25°C\\ T_{final}=20°C\\ m=2.5g\\ Q=-12cal\end{array}$

Substitute the values in the above formula for heat absorbed or released,

$\begin{array}{l}Q=m\times c\times \Delta T\\ c=\frac{Q}{m\times \Delta T}\\ c=\frac{-12cal}{2.5g\times \left(20°C-25°C\right)}=\frac{-12cal}{2.5g\times \left(-5°C\right)}\\ c=\frac{-12\times 4.186J}{2.5g\times \left(-5°C\right)}=4.019\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$\left(g°C\right)$}\right.\end{array}$

Explanation of Solution

The formula to calculate the heat absorbed or released by a substance is rearranged to calculate specific heat. The given value of heat released is converted into joules from calories. The mass, heat released in joules, and the change in temperature are substituted in the rearranged formula to calculate the value of specific heat.

Conclusion

The specific heat, c, of the unknown substance is,

$\begin{array}{l}Q=m\times c\times \Delta T\\ c=\frac{Q}{m\times \Delta T}\\ c=4.019\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$\left(g°C\right)$}\right.\end{array}$