Question

1 g of ice at 0C is mixed with 1g of steam at 100C. After thermal equilibrium is attained the temperature

of the mixture

(A) 1C (B) 50C

(C) 81C (D) 100C​

Answer 1

Answer:

c is the correct answer

Answer 2

check-circle

Text Solution

Answer :

C

Solution :

Heat requiired by 1 g ice at

C to ment into 1 g water at

C. <br>

= mL (Latent heat of fusion L =

<br>

<br> Heat required by 1g of water of

C to boil at

C,

(specific heat of water c = 1cal/

C) <br>

<br> = 100 cal <br> Thus, total heat required by 1g of ice to reach a temperature of

C, <br> Q =

= 80 + 100 =180 cal <br> Heat available with 1 g of ice to condense into 1g of water at

C, <br>

(Latent of vaporisation

= 536 cal/g) <br>

cal <br> = 536 cal <br> Obvioulsy, the whole steam will not be condensed and ice will attain temperature of

C. Thus, the temperature of mixture is