Chapter U5.6, Problem 9E

Interpretation Introduction

Interpretation:The conservation of energy is to be shown in the two given processes.

Concept introduction: According to law, of conservation of energy, energy is always conserved. Energy cannot be created or destroyed in a reaction.

The formula to calculate heat transfer is,

  $H=mc\Delta T$

Where H is heat transfer, m is mass, c is specific heat capacity and $\Delta T$ is the change of temperature.

The formula to calculate heat transfer using the mass and heat of fusion is,

  $H=mL$

Where H is heat transfer and L is the heat of fusion.

Answer to Problem 9E

Conservation of energy is shown.

Explanation of Solution

25 g ice melts and then goes to $20°\text{C}$ . The energy required in this process is supplied from the energy releases from the cooling of 100 g water to $20°\text{C}$ . Thus, for energy conservation, we need to find that energy required in process 1 is the same as the energy released from process 2.

PROCESS 1:

Heat of fusion of water (S)=80 cal/g

Mass of sample= 25 g

Heat required to melt can be calculated as,

  $\begin{array}{c}\text{Heat'=}\text{Mass}\times \text{Heat}\text{of}\text{fusion}\end{array}$

    $\begin{array}{c}\text{=25}\text{g}\times \text{80}\text{cal/}\text{g}\end{array}$

    $\begin{array}{c}\text{=2000}\text{cal}\end{array}$

Specific heat of water (c)= ${\text{1cal/g }}^{\text{o}}\text{C}$

Final temperature= $20°\text{C}$

  $\Delta T=20°\text{C}$

Heat transferred can be calculated as,

  $\begin{array}{c}Heat"=mc\Delta T\end{array}$

     $\begin{array}{c}=25\text{g}\times 1\text{g}/\text{cal}°\text{C}\times \text{20}°\text{C}\end{array}$

     $\begin{array}{c}\text{=500}\text{cal}\end{array}$

  $\begin{array}{c}\text{Total}\text{heat=}\text{Heat'}\text{+}\text{Heat"}\end{array}$

      $\begin{array}{c}\text{=2000}\text{cal}\text{+500}\text{cal}\end{array}$

      $\begin{array}{c}\text{=2500}\text{cal}\end{array}$

PROCESS 2:

Mass (m)=100g

Change in temperature $\left(\Delta T\right)=45-20=25°\text{C}$

Specific heat = $\text{1 cal}/\text{ g}°\text{C}$

$\begin{array}{c}\text{Heat}\text{transferred=}mc\Delta T\end{array}$

       $\begin{array}{c}=100\text{g}\times \text{1}\text{cal/}\text{g}°\text{C}\times \text{25}°\text{C}\end{array}$

       $\begin{array}{c}\text{=2500}\text{cal}\end{array}$

From process (1) and (2), it is clear that energy is conserved.

Conclusion

Energy is always conserved.