Chapter 10.2, Problem 4RQ

Interpretation Introduction

Interpretation:The energy required to heat a mug of water (250 mL) needs to be calculated in calorie and joules.

Concept introduction:

The formula to calculate energy is as follows:

  $Q\text{=}s\times m\times \Delta T$  ...... (1)

Where Q is energy, s is specific heat capacity and $\Delta T$ is change of temperature and m is mass of sample in gram.

Formula to calculate density is as follows:

  $\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 4RQ

The energy required to heat 250 mL of water is 78450 J (in Joule) and 18750 calories (in calorie).

Explanation of Solution

The volume of water is 250 mL and the density of water is $1\text{g}\text{/}\text{mL}$ . Now use the formula of density to calculate the mass of water.

  $\begin{array}{c}\text{Density=}\frac{\text{Mass}}{\text{Volume}}\end{array}$

    $\begin{array}{c}\text{Mass=1}\text{g/mL}\times \text{250}\text{mL}\end{array}$

       $\begin{array}{c}\text{=250}\text{g}\end{array}$

So, the mass of water is 250 g.

  $\begin{array}{c}{T}_1=25°\text{C}\end{array}$

   $\begin{array}{c}{T}_2=100°\text{C}\end{array}$

  $\begin{array}{c}\Delta T=100°\text{C}-\text{25}°\text{C}\end{array}$

     $\begin{array}{c}\text{=75}°\text{C}\end{array}$

The specific heat capacity of water is $4.184\text{J}\text{/}\text{g}°\text{C}$ . Now, put all the values in equation (1) to calculate heat.

  $\begin{array}{c}Q=250\text{g}\times 7\text{5}°\text{C}\times 4.184\text{J}/\text{g}°\text{C}\end{array}$

    $\begin{array}{c}\text{=78450}\text{J}\end{array}$

  $1\text{cal=4}\text{.184}\text{J}$

So, the conversion factor to convert energy from joule to calorie is $\frac{1\text{cal}}{4.184\text{J}}$

  $\begin{array}{c}\text{Energy}\text{in}\text{calorie}=78450\text{J}\times \frac{1\text{cal}}{4.184\text{J}}\end{array}$

          $\begin{array}{c}=18750\text{cal}\end{array}$

Conclusion

Joule and Calorie both are units of energy.