Chapter 14, Problem 19P

To determine

The specific heat of the sample of substance.

Explanation of Solution

Formula used:

The heat gained by Aluminum is equal to the heat lost by the substance

${{\displaystyle m}}_x{{\displaystyle c}}_x\left({{\displaystyle T}}_{ix}-{{\displaystyle T}}_{eq}\right)={{\displaystyle m}}_{Al}{{\displaystyle c}}_{Al}\left({{\displaystyle T}}_{eq}-{{\displaystyle T}}_{iAL}\right)+{{\displaystyle m}}_w{{\displaystyle c}}_w\left({{\displaystyle T}}_{eq}-{{\displaystyle T}}_{iw}\right)+{{\displaystyle m}}_g{{\displaystyle c}}_g\left({{\displaystyle T}}_{eq}-{{\displaystyle T}}_{ig}\right)$

${{\displaystyle m}}_x$ = mass of substance

${{\displaystyle m}}_{Al}$ = mass of aluminum

${{\displaystyle m}}_w$ = mass of water

${{\displaystyle m}}_g$ = mass of glass

${{\displaystyle T}}_{ix}$ = temperature of substance

${{\displaystyle T}}_{eq}$ = equilibrium temperature.

${{\displaystyle T}}_{eq}$ = temperature of water

${{\displaystyle T}}_{ig}$ = temperature of glass

$\begin{array}{l}{C}_w=\text{specific heat of water}\\ C_{Al}=\text{specific heat of aluminum}\\ C_g=\text{ specific heat of glass} \end{array}$

Calculation:

The heat gained by Aluminum is equal to the heat lost by the substance;

$\begin{array}{l}{{\displaystyle m}}_x{{\displaystyle c}}_x\left({{\displaystyle T}}_{ix}-{{\displaystyle T}}_{eq}\right)={{\displaystyle m}}_{Al}{{\displaystyle c}}_{Al}\left({{\displaystyle T}}_{eq}-{{\displaystyle T}}_{iAL}\right)+{{\displaystyle m}}_w{{\displaystyle c}}_w\left({{\displaystyle T}}_{eq}-{{\displaystyle T}}_{iw}\right)+{{\displaystyle m}}_g{{\displaystyle c}}_g\left({{\displaystyle T}}_{eq}-{{\displaystyle T}}_{ig}\right)\\ {{\displaystyle c}}_x=\frac{(0.105\text{ kg)(900 J/kg }{}^{0}C)+(0.185kg)(4186J/k{g}^{0}C)+(0.017kg)(840J/kg{}^{0}C)]\left(24.5{}^{0}C\right)}{(0.215kg)\left({330}^{0}C-{35}^{0}C\right)}\\ {{\displaystyle c}}_x=3.09\times {10}^2\text{ J/kg}{}^{0}C\end{array}$