Chapter 15, Problem 87A

How much heat is evolved when 1255 g of water condensesto a liquid at 100°C?

Interpretation Introduction

Interpretation:

Heat evolved when 1255g of water condenses to a liquid at $\text{100°C}$ is to be calculated.

Concept introduction:

Molar heat of condensation ${\text{(ΔH}}_{\text{cond}}^{}\text{)}$ is the amount of heat released when 1mol of vapor condenses.

${\text{ΔH}}_{\text{vap}}^{}{\text{=-ΔH}}_{\text{cond}}^{}$

Answer to Problem 87A

2837kJ heat will be released during the process.

Explanation of Solution

Let us first convert grams of water to moles of water(molar mass of water is $\text{18g/mol}$ )

$\text{Moles=}\frac{\text{Grams}\text{of}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{compound}}$

For, 1255g of water

${\text{=1255gH}}_{\text{2}}^{}\text{O}\text{.}\frac{\text{1mol}}{\text{18g}}$

$\text{=69}{\text{.7molH}}_{\text{2}}^{}\text{O}$

Since, ${\text{ΔH}}_{\text{vap}}^{}{\text{=-ΔH}}_{\text{cond}}^{}$

The molar heat of vaporization ${\text{(ΔH}}_{\text{vap}}^{})$ of ${\text{H}}_{\text{2}}^{}\text{O}$ is ${\text{ΔH}}_{\text{vap}}^{}\text{=}\text{40}\text{.7kJ/mol}$

Then, ${\text{ΔH}}_{\text{cond}}^{}\text{=}\text{-40}\text{.7kJ/mol}$

$\text{ΔH=69}{\text{.7molH}}_{\text{2}}^{}\text{O}\text{.}\frac{\text{-40}\text{.7kJ}}{\text{1mol}{\text{H}}_{\text{2}}^{}\text{O}}$

$\text{=}\text{-2837kJ}$

Conclusion

The negative sign indicates that the heat is given off..