Question

the amount of heat released when one gram of steam condensed to boiling water is
a)80cal
b)540cal
c)260cal
d)340cal​

Answer 1

Answer:

ANSWER

Let, L

f

= Latent Heat of fusion,

L

v

= Latent Heat of Vaporization,

s= specific heat, and

dθ= change in Temperature

To melt 100g ice at 273K, heat required =mL

f

=8000cal

To convert 100g water at 273K to 100g water at 373K:

Heat required =msdθ=100×1×100=10000cal

To convert 100g water at 373K to steam at 373K, heat required = mL

v

=54000cal

Thus, total heat required to convert 100g ice at 273K to 100g water at 373K is 8000+10000=18000cal

But to convert the same quantity of ice to steam at 373K, heat required will be 72000cal

The heat supplied is 22320cal, therefore, all of ice is not converted to steam. But all of it is necessarily converted to water at 373K first and then some of this water is converted to steam. The additional heat available after all ice has become water at 373K is 22320−18000=4320cal.

Thus if M grams of water is converted to steam then, ML

v

=4320⇒M=

540

4320

=8g.

So, the final mixture is 92g water and 8g steam in equilibrium at 373K.

Answer 2

Answer:

I don't know the full answer

can I give half answer