Respuesta :
The minimum heat required is 129.7KJ.
How to find the minimum heat?
Given
Mass of water (m) = 50g
Initial temperature [tex]T_{i}[/tex] = 20°C
Final temperature [tex]T_{f}[/tex] = 100°C
Specific heat of water =4190 J/kg-k
The heat of vaporization (L) = 22.6 × 10 5 j/kg .
The heat required = Heat required to raise water temperature from 20 to 100 + Heat to vapourize water thoroughly.
Q=mc([tex]T_{f} - T_{i}[/tex]) + mL
Q=0.05 * 4190 (100-20) + 0.05 * 22.6 × [tex]10^{5}[/tex]
Q=(0.16760 + 1.13)× [tex]10^{5}[/tex]
Q=1.2976 × [tex]10^{5}[/tex]
= 129.7KJ
To learn more about heat, refer
https://brainly.com/question/21927103
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