Chapter 12.2, Problem 32SSC

To determine

The energy needed to heat mercury from $10°\text{C}$ to its boiling point and vaporize it.

Answer to Problem 32SSC

  $\text{354}\text{.6 kJ}$

Explanation of Solution

Given Specific heat capacity of mercury, $C=140\text{J/kg}\cdot °\text{C}$

Latent heat of vaporization, $H_v=3.06\times {10}^5\text{J/kg}$

Final temperature is $T_f=357°\text{C}$

Initial temperature, $T_i=10°\text{C}$

Mass of mercury, $m=1\text{kg}$

Formula Used:

Energy needed to heat a material is $Q=m\times C\times (T_f-T_i)$

To vaporize a material at boiling point,required heat is: $Q\text{'}=m{H}_v$

Total energy needed is $m\times C\times (T_f-T_i)+m{H}_v$

Calculations:

Substitute the values and solve:

Energy required is

  $\begin{array}{c}{Q}_T=\text{1kg}\text{×140J/kg}\cdot °\text{C×(357}°\text{C-10}°\text{C)+1}\text{kg}\times \text{3}{\text{.06×10}}^{\text{5}}\text{J/kg}\\ \text{=354580 J}\\ \approx \text{354}\text{.6 kJ}\end{array}$

Conclusion:

The energy needed to heat $1kg$ of mercury from $10C$ to its boiling point and vaporize it is $\text{354}\text{.6 kJ}$