Question

: How much quantity of heat is required?

m = 120 litres​

Answer 1

Answer:

One liter of water is about 1 Kg or 1000g, giving Q = m x (delta T) x Specific heat of water = 1000g x 80C x 1 cal/g C= 80,000 cal or 80 Kcals. Technically, 80 kcals that is the amount of heat needed to get the liquid water to 100C and anything extra would go into vaporizing the water into steam.

Step-by-step explanation:

Weight of one litre of water is about

1Kg = 1000g

Q = m x (delta T) x Specific heat of water

1000g x 80C x 1 cal/g C= 80,000 cal

80,000 cal= 80Kcal