Answer to Question #140280 in General Chemistry for Katherine

Question

How much energy (heat) is required to convert 35.8 g of H2O(l) at 42.0°C to H2O(g) at 105.0°C?

Specific heat water = 4.18 J/g/oC

Specific heat steam = 1.84 J/g/oC

ΔHvap (water) = 40.7 kJ/mol

Answer 1

Solution:

Three calculations, then add the results of those calculations for the final answer.

1) Energy required to warm the water from 42.0 C to 100.0 C

2) Energy required to boil the water at 100.0 C

3) Energy required to warm the steam from 100.0 C to 105.0 C

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$Q_{1}=(35.8g)(58^{o} C)(4.184( \tfrac{J}{g*C} ))=8687.65J = 8.68kJ$

$Q_{2}=( \dfrac{35.8g}{18( \tfrac{g}{mol} )} )(40.7( \tfrac{kJ}{mol} )=80.94 kJ$

$Q_{3}=(35.8g)(5^{o}C)(1.84 ( \tfrac{J}{g*^{o}C} ))=329.36J=0.329kJ$

$Q_{total}=Q_1+Q_2+Q_3=8.68+80.94+0.329=90kJ$

Answer:

$Q_{total} = 90kJ$