Question

considering a latent heat of ice=80cal/gm and latent heat of steam =540cal/gm the amount of heat required in calories to convert 5gms of ice at 0 degree celsius to steam at 100 degree celsius will be

Answer 1

Latent Heat of Fusion = Lf = 80 cal/g
Latent Heat of Vaporisation = Lv = 540 cal/g
Specific heat of water = c = 1 cal/g

Mass of ice = m = 5 g

→ First, we need to convert ice at 0°C into water at 0°C. For this, we need to supply Latent Heat of Fusion.

Q1 = m×Lf
So, Q1 = 5×80 = 400 cal

→ Now, we need to convert water at 0°C into water at 100°C.

For this, we need to consider specific heat. The amount of heat energy required to raise the temperature of a 1 gram substance by 1°C is called specific heat.

Specific heat of water is 1 cal/g
It's formula is:
c = (Q)/(m×∆T)
where ∆T is change in temperature.

Thus Q = m×c×∆T
Here, ∆T = (100°C - 0°C) = 100 °C

So, Q2 = m×c×∆T
So, Q2 = 5×1×100
So, Q2 = 500 cal


→ Finally, we need to convert water at 100°C into steam at 100°C. For this, we need to supply Latent Heat of Vaporisation.

Thus, Q3 = m×Lv
So, Q3 = 5×540
So, Q3 = 2700 cal

→ Thus, now the total heat is:

Q = Q1 + Q2 + Q3
So, Q = 400 + 500 + 2700 cal
So, Q = 3600 cal = 3.6 kcal

Thus, the amount of heat required to convert 5 grams of ice at 0°C into steam at 100°C will be 3600 calories.