What is the heat required in kilocalories to convert 2 kg of ice at 0°c completely into steam at 100°c? a. 80 calories b. 1440 calories c. 4186 calories d. 540 calories

Q = mL 

We need to know the latent heat of fusion, L, for 0 degrees.

L = 336 kJ/kg, m = 2kg

where L is the latent heat for vaporization

Q = 2 * 336 = 672 kJ

For conversion between 0 and 100 degree Celsius

Q = mcθ,   specific heat capacity = 4.2 kJ/kgK

Q = 2*4.2* (100 - 0) = 840 kJ

For conversion to steam at 100 degrees Celsius

Q = mL ,  L = latent heat of vaporization = 2256 kj/kg

Q = 2 * 2256 = 4512 kJ

Total heat = 672 + 840 + 4512 = 6024 kJ

= 6 024 000 J

But 1 calorie = 4.2 J

Therefore 6 024 000 J will be:           6 024 000/4.2 ≈ 1 434 286 Calories

≈ 1 434 kCalories