# Question 2a1c5

Apr 28, 2015

Since you didn't provide an initial temperature for the water, I'll show you how to calculate the energy needed to boil water at ${100}^{\circ} \text{C}$.

The enthalpy of vaporization, or latent heat of vaporization, is the heat you need to provide to a mole of water to convert it from liquid at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$.

To boil 1 mole of water at ${100}^{\circ} \text{C}$ you need to provide 40.67 kJ of heat - this amount of heat will cause one mole of water to undergo a phase change.

Determine how many moles of water you're dealing with by using water's molar mass

150cancel("g") * "1 mole water"/(18.015cancel("g")) = "8.326 moles of water"

The heat needed to boil this much water will be

$q = n \cdot \Delta {H}_{\text{vap}}$

q = 8.326cancel("moles") * 40.67"kJ"/cancel("mole") = "338.62 kJ"#

Rounded to two sig figs, the number of sig figs given for 150 g, the answer will be

$q = \textcolor{g r e e n}{\text{+340 kJ}}$