Answer:
7.7 MJ
Explanation:
Let water specific heat be c = 0.004186 J/kgC and specific latent heat of vaporization be L = 2264705 J/kg
There would be 2 kinds of heat to achieve this:
- Heat to change water temperature from 78C to boiling point:
[tex]H_1 = mc\Delta t = 3.4 * 0.004186 * (100 - 78) = 0.313 J[/tex]
- Heat to turn liquid water into steam:
[tex]H_2 = mL = 3.4 * 2264705 \approx 7.7 \times 10^6J[/tex] or 7.7 MJ
So the total heat it would take is [tex]H_1 + H_2 = 7.7 \times 10^6 + 0.313 \approx 7.7 MJ[/tex]
