Respuesta :
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
Answer : The heat required will be, 20.35 KJ
Explanation : Given,
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]H_2[/tex]
[tex]\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{9g}{18g/mole}=0.5mole[/tex]
Now we have to calculate the heat produced.
Formula used :
[tex]Q=n\times \Delta H_{vap}[/tex]
where,
Q = heat required = ?
n = moles of water = 0.5 mole
[tex]\Delta H_{vap}[/tex] = 40.7 KJ/mole
Now put all the given values in this formula, we get the heat required.
[tex]Q=0.5mole\times 40.7KJ/mole=20.35KJ[/tex]
Therefore, the heat required will be, 20.35 KJ
