Question

10 gram of steam passes over a ice block at zero degree celsius what amount of ice melt

Answer 1

Answer:

For thermal equilibrium the final temperature of steam must be equal to ice i.e. zero.

The steam will go phase transition from gas to liquid, heat released is given as

Q=mL where L is latent heat of vaporization and m is mass of substance

For water L

vaporization

=540cal/g

Q

1

=10∗540=5400cal

Boiling water will change it's temperature from 100

∘

C to 0

∘

C

heat released is given as

Q=mCΔTcal/g where C is specific heat capacity and ΔT is temperature difference

Q

2

=10×1×(100−0);C

water

=1cal/g

total heat transferred=Q

1

+Q

2

=6400cal

This amount of heat will be used for melting the ice

6400=mL

fusion

=m80cal/g latent heat of fusion for water =80cal/g

⇒m=80g

please mark this as the brainiest answer :(