Chapter 10, Problem 59A

Interpretation Introduction

Interpretation: The raise in temperature of each sample needs to be determined.

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 59A

Amount of temperature that rises in liquid water is $23.9°\text{C}$ .

Amount of temperature that rises in solid water is $49.3°\text{C}$ .

Amount of temperature that rises in water vapor is $50°\text{C}$ .

Amount of temperature that rises in alumna is $112.36°\text{C}$ .

Amount of temperature that rises in iron vapor is $222.22°\text{C}$ .

Amount of temperature that rises in mercury is $714.29°\text{C}$ .

Amount of temperature that rises in carbon is $140.85°\text{C}$ .

Amount of temperature that rises in silver is $416.67°\text{C}$ .

Amount of temperature that rises in gold is $769.23°\text{C}$ .

Amount of temperature that rises in copper is $259.74°\text{C}$ .

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 10.0g, heat applied is 1.00 kJ.

Conversion of heat is as follows:

  $\begin{array}{c}\text{1}\cancel{\text{kJ}}\text{×}\frac{\text{1000J}}{\text{1}\cancel{\text{kJ}}}\text{=1000J}\\ {\text{=1×10}}^{\text{3}}\text{J}\end{array}$

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:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=4}\text{.184}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{\text{4}\text{.184}\cancel{\text{J}}/°\text{C}\times 10}\\ =23.9°\text{C}\end{array}$

Thus rises in temperature is $23.9°\text{C}$ .

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

To rise in temperature that is required by solid water sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=2}\text{.03}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{2.03\cancel{\text{J}}/°\text{C}\times 10}\\ =49.26°\text{C}\\ \approx 49.3°\text{C}\end{array}$

Thus rises in temperature is $49.3°\text{C}$ .

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

To rises in temperature that is required by water vapor sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=2}\text{.0}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{2.0\cancel{\text{J}}/°\text{C}\times 10}\\ =50°\text{C}\end{array}$

Thus rises in temperature is $50°\text{C}$ .

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

To rises in temperature that is required by alumina sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.89}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.89\cancel{\text{J}}/°\text{C}\times 10}\\ =112.36°\text{C}\end{array}$

Thus rises in temperature is $112.36°\text{C}$ .

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

To rises in temperature that is required by iron sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.45}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.45\cancel{\text{J}}/°\text{C}\times 10}\\ =222.22°\text{C}\end{array}$

Thus rises in temperature is $222.22°\text{C}$ .

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

To rises in temperature that is required by mercury sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.14}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.14\cancel{\text{J}}/°\text{C}\times 10}\\ =714.29°\text{C}\end{array}$

Thus rises in temperature is $714.29°\text{C}$ .

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

To rises in temperature that is required by carbon sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.71}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.71\cancel{\text{J}}/°\text{C}\times 10}\\ =140.85°\text{C}\end{array}$

Thus rises in temperature is $140.85°\text{C}$ .

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

To rises in temperature that is required by silver sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.24}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.24\cancel{\text{J}}/°\text{C}\times 10}\\ =416.67°\text{C}\end{array}$

Thus rises in temperature is $416.67°\text{C}$ .

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

To rises in temperature that is required by gold sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.13}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.13\cancel{\text{J}}/°\text{C}\times 10}\\ =769.23°\text{C}\end{array}$

Thus rises in temperature is $769.23°\text{C}$ .

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

To rises in temperature that is required by copper sample in equation (1) is as follows:

  $\begin{array}{c}\text{1}\times {\text{10}}^{\text{3}}\text{J=0}\text{.385}\frac{\text{J}}{\cancel{\text{g}}°\text{C}}\text{×10}\cancel{\text{g}}\text{×}\Delta \text{T}\\ \Delta \text{T}=\frac{\text{1}\times {\text{10}}^{\text{3}}\cancel{\text{J}}}{0.385\cancel{\text{J}}/°\text{C}\times 10}\\ =259.74°\text{C}\end{array}$

Thus rises in temperature is $259.74°\text{C}$ .

Conclusion

Amount of temperature that rises in liquid water, solid water and water vapor, alumina, iron, mercury, carbon, silver, gold and copper are $23.9°\text{C}$ , $49.3°\text{C}$ , $50°\text{C}$ , $112.36°\text{C}$ , $222.22°\text{C}$ , $714.29°\text{C}$ , $140.85°\text{C}$ , $416.67°\text{C}$ , $769.23°\text{C}$ , $259.74°\text{C}$ respectively.