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* c_(p,l)* (T_(final)-T_(initial))]+n* \Delta H_(vap)+[m* c_(p,g)* (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}}=(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* 4.184J/gK* (100-50)^oC]+6.11mole* 40670J/mole+[110.0g* 1.84J/gK* (110-100)^oC]`

`\Delta H=273529.7J`

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