Question

60 g of ice at 0°C is mixed with 60 g of steam
at 100°C. At thermal eqilibrium, the mixture
contains

1) 80 g of water and 40 g of steam at 100°C
2) 120 g of water 90°C
3) 120 g of water at 100°C
4) 40 g of steam and 80 g of water at 0°C​

Answer 1

Explanation:

Top answer

Heat energy required to melt 60 g of ice = 60 × 80 = 4800 Cal .Heat energy required to raise the temprature of water at 0 degree centigrade to 100 ...