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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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100 g of ice at $\;0^{0}C\;$ is mixed with 100 g of water at $\;20^{0}C\;$ . Latent heat of fusion of fusion is 80 cal/g & specific heat of water is $\;1 cal /g^{0}C\;$ . The final temperature of mixture =

$(a)\;20^{0}C\qquad(b)\;10^{0}C\qquad(c)\;5^{0}C\qquad(d)\;0^{0}C$

Can you answer this question?
 
 

1 Answer

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Answer : (d) $\;0^{0}C$
Explanation :
Heat required to melt ice = $100 \times 80 = 8 \;kcal$
Max . Heat that can be lost by water
$=100 \times 1 \times (20-0)$
$= 2 \;kcal$
Therefore , whole of the ice will not melt . Therefore , the temperature remains $\;0^{0}C$
answered Mar 12, 2014 by yamini.v
 

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