Chapter 10.2, Problem 3RQ

Interpretation Introduction

Interpretation: The given table needs to be completed.

  

Concept introduction: The formula to calculate energy is,

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

Answer to Problem 3RQ

The completed table is,

Mass of water	Change in temperature	Energy (cal)	Energy (J)
50.0 g	$32.8°\text{C}$	6861.76 J	1640 cal
0.488 g	$21.4°\text{C}$	10.44 cal	43.7 J
24.0 g	$130.86°\text{C}$	$3.14\times {10}^3\text{cal}$	$\text{13}\text{.14}\times {10}^3\text{J}$

Explanation of Solution

For case 1:

  $\text{Specific}\text{heat}\text{of}\text{water=4}\text{.184}\text{J}/g°\text{C}$

Mass of water is 50 g and temperature difference is $32.8°\text{C}$ .

Put all the values in equation (1).

  $\begin{array}{c}Q\text{=}s\times m\times \Delta T\\ =4.184\text{J}/g°\text{C}\times 50\text{g}\times \text{32}\text{.8}°\text{C}\\ \text{=6861}\text{.76}\text{J}\end{array}$

So, energy in joule is 6861.76 J

  $1\text{cal=4}\text{.184}\text{J}$

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

  $\begin{array}{c}\text{Energy}\text{in}\text{calorie}=6861.76\text{J}\times \frac{1\text{cal}}{4.184\text{J}}\\ =1640\text{cal}\end{array}$

So, energy in calorie is 1640 cal.

For case 2:

Change in temperature is $21.4°\text{C}$ and energy in joule is 43.7 J. So, put all the values in equation (1) to calculate mass of water.

  $\begin{array}{c}43.7\text{J=}m\times 21.4°\text{C}\times \text{4}\text{.184}\text{J/g}°\text{C}\\ m\text{=}\frac{43.7\text{J}}{\text{4}\text{.184}\text{J/g}°\text{C}\times 21.4°\text{C}}\\ =0.488g\end{array}$

So, mass of water is 0.488 g.

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

  $\begin{array}{c}\text{Energy}\text{in}\text{calorie}=43.7\text{J}\times \frac{1\text{cal}}{4.184\text{J}}\\ =10.44\text{cal}\end{array}$

So, energy in calorie is 10.44 cal.

For case 3:

Mass of water is 24.0 g and energy in calorie is $3.14\times {10}^3\text{cal}$ .

Conversion factor to convert energy from calorie to joule is $\frac{4.184\text{J}}{1\text{cal}}$

So,

  $\begin{array}{c}\text{Energy}\text{in}\text{Joule=}\frac{4.184\text{J}}{1\text{cal}}\times 3.14\times {10}^3\text{cal}\\ \text{=13}\text{.14}\times {10}^3\text{J}\end{array}$

So, energy in Joule is $\text{13}\text{.14}\times {10}^3\text{J}$

Now use equation (1) to calculate change in temperature.

  $\begin{array}{c}\text{13}\text{.14}\times {10}^3\text{J=24}\text{g}\times \Delta T\times 4.184\text{J/}\text{g}°\text{C}\\ \Delta T=\frac{\text{13}\text{.14}\times {10}^3\text{J}}{\text{24}\text{g}\times 4.184\text{J/}\text{g}°\text{C}}\\ =130.86°\text{C}\end{array}$

So, temperature difference is $130.86°\text{C}$ .

Conclusion

Joule and Calorie both are units of energy.