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

Thermodynamics

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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 $\;0^{0}C$

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