Chapter 5, Problem 45R

To determine

The change in temperature when graphite of $33\text{ g}$ absorbs $350\text{ J}$ of thermal energy.

Answer to Problem 45R

  $\Delta T=14.73°\text{C}$

Explanation of Solution

Given:

Thermal energy, $Q=350\text{J}$

Specific heat of graphite, $C=0.72\text{J/g}°\text{C}$

Mass of the sample, $m=33\text{ g}$

Formula used:

Thermal energy can be calculated as:

  $Q=mc\left(\Delta T\right)$

Where,

  $m$ = mass

  $c$ = Heat capacity of graphite

  $\left(\Delta T\right)$ = change in temperature

Calculation:

  $\begin{array}{l}Q=mC\Delta T\\ 350=33\left(0.72\right)\Delta T\\ \Delta T=14.73°\text{C}\end{array}$

Conclusion:

Change in temperature is $\Delta T=14.73°\text{C}$