Question

5 g of ice at 0 c is mixed with 5g of steam at 100 c what is final temp

Answer 1

Heat energy required for

5

g

of water at

0

∘

C

to get converted to water at

100

∘

C

is latent heat required + heat required to change its temperature by

100

∘

C

=

(

80

⋅

5

)

+

(

5

⋅

1

⋅

100

)

=

900

calories.

Now,heat liberated by

5

g

of steam at

100

∘

C

to get converted to water at

100

∘

C

is

5

⋅

537

=

2685

calories

So,heat energy is enough for

5

g

of ice to get converted to

5

g

of water at

100

∘

C

So,only

900

calories of heat energy will be liberated by steam,so amount of steam that will be converted to water at the same temperature is

900

537

=

1.66

g

So,the final temperature of the mixture will be

100

∘

C

in which

5

−

1.66

=

3.34

g

of steam and

5

+

1.66

=

6.66

g

of water will coexist.

I hope this will help you

If not then comment me

Answer 2

hlooo mate ur ans....!!!!