Question

Heat is added to 110.0 grams of liquid of water at 50.00°C to produce water vapor. The vapor is collected and heated to a temperature of 110.0°C. How many joules of heat energy were required in total for this process?

Answer 1

Answer : The heat energy required in total for this process was 273529.7 joules.

Solution :

The conversions involved in this process are :

$(1):H_2O(l)(50^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(110^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of water = 110.0 g

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of water vapor = $1.84J/g^oC$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{110.0g}{18g/mole}=6.11mole$

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[110.0g\times 4.184J/gK\times (100-50)^oC]+6.11mole\times 40670J/mole+[110.0g\times 1.84J/gK\times (110-100)^oC]$

$\Delta H=273529.7J$

Therefore, the heat energy required in total for this process was 273529.7 joules.