Chapter 10, Problem 58A

Interpretation Introduction

Interpretation: The quantity of heat required to heat 150g of the given substances by $11.2°\text{C}$ to be calculated. 

Concept introduction: Due to varying temperature flow of energy is termed as heat. It is depend on the pathway. Heat is expressed in energy units, typically calories and joules.

Answer to Problem 58A

Amount of heat required in liquid water is $\text{7}\text{.03}\times {\text{10}}^{\text{3}}\text{J}$ .

Amount of heat required in solid water is $\text{3}\text{.41}\times {\text{10}}^{\text{3}}\text{J}$ .

Amount of heat required in water vapor is $\text{3}\text{.4}\times {\text{10}}^{\text{3}}\text{J}$ .

Amount of heat required in alumna is $\text{1}\text{.5}\times {\text{10}}^{\text{3}}\text{J}$ .

Amount of heat required in iron vapor is $7.6\times {\text{10}}^{\text{3}}\text{J}$ .

Amount of heat required in mercury is $2.4\times {\text{10}}^2\text{J}$ .

Amount of heat required in carbon is $1.2\times {\text{10}}^2\text{J}$ .

Amount of heat required in silver is $4.0\times {\text{10}}^2\text{J}$ .

Amount of heat required in gold is $2.2\times {\text{10}}^2\text{J}$ .

Amount of heat required in copper is $6.47\times {\text{10}}^2\text{J}$ .

Explanation of Solution

The expression for heat is as follows:

  $\text{Q=s×m×ΔT}$ …… (1)

Here Q is heat capacity, s is specific heat capacity, m is mass of the sample, and $\text{ΔT}$ is change in temperature.

Given mass of the sample is 150g, temperature change is $11.2°\text{C}$ .

Heat capacity of liquid water is $\text{4}\text{.184}\frac{\text{J}}{\text{g°C}}$ .

Put these values in equation (1) is as follows:

  $\begin{array}{c}\text{Q=4}\text{.184}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =7029.12\text{J}\\ \approx \text{7}\text{.03}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Thus amount of heat required is $\text{7}\text{.03}\times {\text{10}}^{\text{3}}\text{J}$ .

Heat capacity of solid water is $\text{2}\text{.03}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by solid water sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=2}\text{.03}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =3410.4\text{J}\\ \approx \text{3}\text{.41}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Thus amount of heat required is $\text{3}\text{.41}\times {\text{10}}^{\text{3}}\text{J}$ .

Heat capacity of water vapor is $\text{2}\text{.0}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by water vapor sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=2}\text{.0}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =3360\text{J}\\ \approx \text{3}\text{.4}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Thus amount of heat required is $\text{3}\text{.4}\times {\text{10}}^{\text{3}}\text{J}$ .

Heat capacity of aluminum is $\text{0}\text{.89}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by alumina sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=8}\text{.9}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =14952\text{J}\\ \approx 1.5\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Thus amount of heat required is $\text{1}\text{.5}\times {\text{10}}^{\text{3}}\text{J}$ .

Heat capacity of iron is $\text{0}\text{.45}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by iron sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=0}\text{.45}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =756\text{J}\\ \approx 7.6\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Thus amount of heat required is $7.6\times {\text{10}}^{\text{3}}\text{J}$ .

Heat capacity of mercury is $\text{0}\text{.14}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by mercury sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=0}\text{.14}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =235.2\text{J}\\ \approx 2.4\times {\text{10}}^2\text{J}\end{array}$

Thus amount of heat required is $2.4\times {\text{10}}^2\text{J}$ .

Heat capacity of carbon is $\text{0}\text{.71}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by carbon sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=0}\text{.71}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =1192.8\text{J}\\ \approx 1.2\times {\text{10}}^3\text{J}\end{array}$

Thus amount of heat required is $1.2\times {\text{10}}^2\text{J}$ .

Heat capacity of silver is $\text{0}\text{.24}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by silver sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=0}\text{.24}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =403.2\text{J}\\ \approx 4.0\times {\text{10}}^2\text{J}\end{array}$

Thus amount of heat required is $4.0\times {\text{10}}^2\text{J}$ .

Heat capacity of gold is $\text{0}\text{.13}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by gold sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=0}\text{.13}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =218.4\text{J}\\ \approx 2.2\times {\text{10}}^2\text{J}\end{array}$

Thus amount of heat required is $2.2\times {\text{10}}^2\text{J}$ .

Heat capacity of copper is $\text{0}\text{.385}\frac{\text{J}}{\text{g°C}}$ .

To calculate heat energy that is required by copper sample in equation (1) is as follows:

  $\begin{array}{c}\text{Q=0}\text{.385}\frac{\text{J}}{\cancel{\text{g}}\cancel{\text{°C}}}\text{×150}\cancel{\text{g}}\text{×11}\text{.2}\cancel{\text{°C}}\\ =646.8\text{J}\\ \approx 6.47\times {\text{10}}^2\text{J}\end{array}$

Thus amount of heat required is $6.47\times {\text{10}}^2\text{J}$ .

Conclusion

Amount of heat required in liquid water, solid water and water vapor, alumina, iron, mercury, carbon, silver, gold and copper are respectively $\text{7}\text{.03}\times {\text{10}}^{\text{3}}\text{J}$ , $\text{3}\text{.41}\times {\text{10}}^{\text{3}}\text{J}$ , $\text{3}\text{.4}\times {\text{10}}^{\text{3}}\text{J}$ , $\text{1}\text{.5}\times {\text{10}}^{\text{3}}\text{J}$ , $7.6\times {\text{10}}^{\text{3}}\text{J}$ , $2.4\times {\text{10}}^2\text{J}$ , $1.2\times {\text{10}}^2\text{J}$ , $4.0\times {\text{10}}^2\text{J}$ , $2.2\times {\text{10}}^2\text{J}$ and $6.47\times {\text{10}}^2\text{J}$ .