Question

C) 100 g ice at 0°C is
converted into steam
at 100 °C. Find
total heat
required (in kcal)
(L, = 80 cal/g,
sv = 1 cal/g°C,
L = 540 cal/g)
. 1a​

Answer 1

Answer:

72 Kcal

Explanation:

1. First for converting 100 g ice at 0 degree C into 100 g water at 0 degree C heat required = 100 × 80 cal = 8000 cal
2. Then for converting 100 g of water at 0 degree C into 100 g of water at 100 degrees C heat required = 100 × 1 × 100 cal = 10000 cal
3. Finally for converting 100 g of water at 100 degrees C into 100 g of steam at 100 degrees C heat required = 100 × 540 cal = 54000 cal
4. So Total heat required = (8000 + 10000 + 54000) cal = 72000 cal = 72 Kcal.