1 g of ice at 0C is mixed with 1g of steam at 100C. After thermal equilibrium is attained the temperature
of the mixture
(A) 1C (B) 50C
(C) 81C (D) 100C
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1 g of ice at 0C is mixed with 1g of steam at 100C. After thermal equilibrium is attained the temperature
of the mixture
(A) 1C (B) 50C
(C) 81C (D) 100C
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Verified answer
Answer:
c is the correct answer
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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