Question

The heat required to completely evaporate $18$ grams of water at ${98}^{\circ }C$ :

$(H_v=2259J/g$ and $c=4.18J/gK$)

• A. $40,812joules$
• B. $40,512joules$
• C. $40,666joules$
• D. $150joules$
• E. $6.12\times {10}^{6}joules$

Answer 1

Answer: (A)

$40,812joules$

For complete evaporation, we have to heat the water from ${98}^{o}C$ to ${100.0}^{o}C$ and then vaporize it at ${100.0}^{o}C$.
$Q=(\text{mass}\times \text{specific heat}\times \Delta T)+(\text{mass}\times \text{Heat of Vaporization})$
$Q=\left(m\right)\times \left(c\right)\times (100.0-98.0)+\left(m\right)\times \left(H_v\right)$
$Q=\left(18\right)\times \left(4.18\right)\times (100.0-98.0)+\left(18\right)\times \left(2259\right)$
$Q=150+40662=40812J$.