Question

16 g of ice is converted to steam at 100 degree Celsius . find the amount of heat required.

( sp heat water=1cal/g degree Celsius
(latent heat at fusion =80cal/g
(latent heat of vapour =540cal/g​

Answer 1

Answer:

Amount of heat required = 11.52 Kcal

Answer 2

The amount of heat required is 11.52 kcal

Explanation:

1. The amount of heat required for conversion of 16g ice at 0°C to 16g water at 0°C is mass of ice * latent heat of fusion.
2. Thus mL = 16*80.
3. Q₁ = 1280 cal.
4. Now, 16g water at 0° C will be converted into 16g water at 100 °C.
5. Q₂ = mcΔT = 16*1*(100-0).
6. Q₂ = 1600 cal.
7. Further, 16g water at 100 °C will be converted into 16g steam at 100 °C.
8. Thus, Q₃ = mL, which is in turn equal to 16*540.
9. Q₃ = 8640 cal.
10. Total heat required for the above process is Q₁+Q₂+Q₃ = 1280+1600+8640.
11. Thus Q = 11,520 cal or 11.52 kcal.