Chapter 10, Problem 53A

Interpretation Introduction

Interpretation: The substance for which the temperature will increase the most and least needs to be determined, if 10.0 J heat is applied to 5.0 g of each sample.

Concept Introduction: Specific heat capacity can be defined as the heat in joules that is required to raise the temperature of 1 g of substance by 1°C. The relation between specific heat capacity and heat required is shown below:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = heat }\\ \text{m = mass }\\ \text{C = specific heat capacity }\\ \text{ΔT= temperature change}\end{array}$

Answer to Problem 53A

Since water has highest specific heat capacity therefore it should have highest temperature change whereas solid gold has least specific heat capacity so it must show highest temperature change.

Explanation of Solution

Heat = 10.0 J

Mass = 5.0 g

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{ΔT=}\frac{\text{q}}{\text{m× C }}\end{array}$

Since heat and mass is same for all the sample therefore the change in temperature must depend on specific heat capacity. The substance with higher value of specific heat capacity must show least temperature change whereas the substance with least value of specific heat capacity must show highest temperature change.

Specific heat capacity for water and gold is 4.184 J/g°C and 0.13 J/g°C respectively.

Since water has highest specific heat capacity therefore it should have highest temperature change whereas solid gold has least specific heat capacity so it must show highest temperature change.

• Specific heat capacity for water = 4.184 J/g°C
Substitute the values to calculate temperature change:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{10 J = 5}\text{.0 g × 4}\text{.184 J/g}°\text{C × ΔT}\\ \text{ΔT}=\frac{\text{10 J}}{\text{5}\text{.0 g × 4}\text{.184 J/g}°\text{C}}\text{ = 0}\text{.478 }°\text{C}\end{array}$

• Specific heat capacity for Gold = 0.13 J/g°C
Substitute the values to calculate temperature change:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{10 J = 5}\text{.0 g × 0}\text{.13 J/g}°\text{C × ΔT}\\ \text{ΔT}=\frac{\text{10 J}}{\text{5}\text{.0 g × 0}\text{.13 J/g}°\text{C}}\text{ = 15}\text{.4}°\text{C}\end{array}$

Conclusion

Thus,

Substance	Temperature change (°C)
Water	0.478
Gold	15.4