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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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$30\; gm$ ice at $\;-20^{0}C\;$ is mixed with $20\; gm$ water at $\;90^{0}C\;$. Find the final composition and temperature of mixture . (Latent Heat of fusion $80\; cal /gm$ )

$(a)\;50 \;gm \;water\;at\;40^{0}C\qquad(b)\;50 \;gm \;water\;at\;-40^{0}C\qquad(c)\;50\; gm\;water\;at\;0^{0}C\qquad(d)\;None$

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1 Answer

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Answer : $None$
Explanation :
ice : 30 gm at $\;-20^{0}C$
water : 20 gm at $\;90^{0}C$
Assume at final stage . whole mixture is water at $\;0^{0}C$
The total heat taken $\;=ms \bigtriangleup T$
$=20 \times 1 \times (90) =1800 \;cal$
Total heat given = $\;ms \bigtriangleup T + mL$
$= 30 \times \large\frac{1}{2} \times (20) + 30 \times 80$
$=300 + 2400 = 2700 \;cal$
We have given more heat so now we will take extra heat from 50 gm water at $\;0^{0}C.$
Extra Heat to be taken = 2700 -1800
$=900\;cal$
This Heat can convert water into ice
$900\;cal = m (80)$
$m=\large\frac{900}{80}=11.2 gm$
only 11.2 gm can be converted into ice
So 11.2 gm ice and 38.8 gm water at $\;0^{0}C$
answered Mar 15, 2014 by yamini.v
 

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