Chapter 10, Problem 57A

Interpretation Introduction

Interpretation: The amount of energy required to heat 150 g of the given substances by 112°C needs to be determined.

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 57A

Substance	Heat required
Water	$\text{7}{\text{.03×10}}^{\text{3}}\text{J}$
Ice	$\text{3}\text{.41×10}\text{J}$
Steam	$\text{3}\text{.4 ×10}\text{J}$
Aluminium	$\text{1}{\text{.5×10}}^{\text{3}}\text{J}$
Iron	$\text{7}{\text{.6×10}}^{\text{3}}\text{ J}$
Mercury	$\text{2}{\text{.4×10}}^{\text{2}}\text{ J}$
Carbon	$\text{1}{\text{.2×10}}^{\text{3}}\text{J}$
Silver	$\text{4}{\text{.0×10}}^{\text{2}}\text{J}$
Gold	$\text{2}{\text{.2×10}}^{\text{2}}\text{J}$
Copper	$\text{6}{\text{.47×10}}^{\text{2}}\text{J}$

Explanation of Solution

Temperature change = 112°C

Mass = 150 g

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

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = 150 g × 4}\text{.184 J/g°C × 112°C }\\ \text{q = 7}{\text{.03×10}}^{\text{3}}\text{J}\end{array}$

• Specific heat capacity for ice = 2.03 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = 150 g × 2}\text{.03 J/g°C × 112°C}\\ \text{q = 3}\text{.41×10}\text{J}\end{array}$

• Specific heat capacity for steam = 2 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{ q = 150 g × 2 J/g°C × 112°C }\\ \text{ q=3}{\text{.4×10}}^{\text{3}}\text{J}\end{array}$

• Specific heat capacity for Aluminium = 0.89 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = 150 g × 0}\text{.89 J/g°C × 112°C}\\ \text{q =1}{\text{.5×10}}^{\text{3}}\text{J}\end{array}$

• Specific heat capacity for iron = 0.45 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{ q = 150 g × 0}\text{.45 J/g°C ×112°C}\\ \text{ q =7}{\text{.6×10}}^{\text{3}}\text{ J}\end{array}$

• Specific heat capacity for liquid mercury = 0.14 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{ q = 150 g × 0}\text{.14 J/g°C × 112°C }\\ \text{q = 2}{\text{.4×10}}^{\text{2}}\text{ J}\end{array}$

• Specific heat capacity for carbon = 0.71 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{ q = 150 g × 0}\text{.71 J/g°C × 112°C}\\ \text{ q =1}{\text{.2×10}}^{\text{3}}\text{J}\end{array}$

• Specific heat capacity for Silver = 0.24 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = 150 g × 0}\text{.24 J/g°C × 112°C }\\ \text{q = 4}{\text{.0×10}}^{\text{2}}\text{J}\end{array}$

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

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = 10}\text{.0 g × 0}\text{.13 J/g°C × 112°C}\\ \text{q = 2}{\text{.2×10}}^{\text{2}}\text{J}\end{array}$

• Specific heat capacity for copper = 0.385 J/g°C
Substitute the values to calculate heat required:

  $\begin{array}{l}\text{q = m× C × ΔT}\\ \text{q = 150 g × 0}\text{.385 J/g°C × 112°C}\\ \text{q = 6}{\text{.47×10}}^{\text{2}}\text{J}\end{array}$

Conclusion

Thus,

Substance	Heat required
Water	$\text{7}{\text{.03×10}}^{\text{3}}\text{J}$
Ice	$\text{3}\text{.41×10}\text{J}$
Steam	$\text{3}\text{.4 ×10}\text{J}$
Aluminium	$\text{1}{\text{.5×10}}^{\text{3}}\text{J}$
Iron	$\text{7}{\text{.6×10}}^{\text{3}}\text{ J}$
Mercury	$\text{2}{\text{.4×10}}^{\text{2}}\text{ J}$
Carbon	$\text{1}{\text{.2×10}}^{\text{3}}\text{J}$
Silver	$\text{4}{\text{.0×10}}^{\text{2}}\text{J}$
Gold	$\text{2}{\text{.2×10}}^{\text{2}}\text{J}$
Copper	$\text{6}{\text{.47×10}}^{\text{2}}\text{J}$