Question

What is the amount of work done to convert 40 g of ice at
-10°C into steam at 100°C? Specific heat of ice= 0.5 cal.g-1.°C-1, latent heat of fusion of ice = 80 cal.-1
latent heat of steam = 540 cal.g-1, J = 4.2 x 10^7 erg. cal-1​

Answer 1

Answer:

1g ice 80 calories 20g ice 80