Question

How much heat is required to convert 9.00 g of water at 100°C into steam at 100°C, if ∆Hvap for water = 40.7 kJ/mol?

Answer 1

Answer : The heat required will be, 20.35 KJ

Explanation : Given,

Molar mass of
`H_2O` = 18 g/mole

First we have to calculate the moles of
`H_2`

`\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=(9g)/(18g/mole)=0.5mole`

Now we have to calculate the heat produced.

Formula used :

`Q=n* \Delta H_(vap)`

where,

Q = heat required = ?

n = moles of water = 0.5 mole

`\Delta H_(vap)` = 40.7 KJ/mole

Now put all the given values in this formula, we get the heat required.

`Q=0.5mole* 40.7KJ/mole=20.35KJ`

Therefore, the heat required will be, 20.35 KJ

Answer 2

There are two types of heat transfers in this process that are as follows:

q1= heat required to warm the water from 100 °C to 100.0 °C.

q2= heat required to vapourize the water into steam at 100 °C.

c=specific heat of water= 4.184 J g⁻¹ °c⁻¹

m=mass of water=9 g

∆Hvap=enthalpy of vapourisation

ΔT=temperature difference

q1=mcΔT=9 × 4.184 J ×0=0 J

q2=mΔHvap= 9 g × 0.0407 J=0.3663 J

q1+q2= (0 + 0.3663) J

= 3.6 × 10⁻¹ J

= 3.6 × 10⁻¹ J