Answer:
80 kcal
Explanation:
The computation of the kilocalories of heat required is shown below:
As we know that
1 liter of water = 1 kg
Therefore the specific heat of water is C = 1 kcal per kg
Now the heat required is
[tex]Q = M\times C \times \Delta T[/tex]
where,
M = 1
C = 1
Delta T = (100 -20)
As we assume that boiling point of water is 100 degrees
Now put this values to the above formula
So, the heat required is
[tex]Q = 1\times 1 \times (100 - 20)[/tex]
= 80 kcal