Chapter 10, Problem 57A

Interpretation Introduction

Interpretation: If $\text{4}\text{.5}\text{J}$ of energy is required to heat $\text{5}\text{.0}\text{g}$ of aluminum sample from $25.0°\text{C}$ to a certain higher temperature, then the amount of energy required to heat $\text{10}\text{.0}\text{g}$ of aluminum over the same temperature interval needs to be determined.

Concept introduction: Specific heat of capacity can be defined as the amount of heat added to a substance to raise the temperature by one degree.

Answer to Problem 57A

The amount of heat required will be $8.9\text{J}$ .

Explanation of Solution

The specific heat capacity for aluminum is $0.89\text{J/g°C}$ .

Let the final temperature be $T_2$ .

Substitute the given values in equation for the determination of final temperature is as follows:

  $\begin{array}{c}Q=s\times m\times \Delta T\\ 4.5\text{J}=0.89\text{J/g°C}\times 5.0\text{g}\times \left(T_2-25.0\right)\text{°C}\\ \left(T_2-25.0\right)\text{°C}=1.0\text{°C}\\ T_2=26.0\text{°C}\end{array}$

Mass of the aluminum sample = $\text{10}\text{.0}\text{g}$ .

Initial temperature $\left(T_1\right)$ = $25.0°\text{C}$

Final temperature $\left(T_2\right)$ = $26.0°\text{C}$

Change in temperature can be calculated as:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =26.0°\text{C}-25.0°\text{C}\\ \text{=1}.0°\text{C}\end{array}$

Now, substitute the values in the equation as:

  $\begin{array}{c}Q=s\times m\times \Delta T\\ =0.89\text{J/g°C}\times 10.0\text{g}\times 1.0\text{°C}\\ =8.9\text{J}\end{array}$

Hence, the amount of heat required will be $8.9\text{J}$ .

Conclusion

The equation for heat is as follows:

  $Q=s\times m\times \Delta T$

Where, Q is the amount of heat, m is the mass of substance, s is the specific heat capacity and $\Delta T$ is the temperature change.