Question

$200\, g$ of ice at $-20^{\circ} C$ is mixed with $500\, g$ of water at $20^{\circ} C$ in an insulating vessel. Final mass of water in vessel is (specific heat of ice $=0.5 \, cal\, g ^{-10}\, C ^{-1}$ )

• A. 700 g
• B. 600 g
• C. 400 g
• D. 200 g

Answer 1

Answer: (B)

Maximum heat supplied by water
$\Delta Q_{1} =500 \times 1 \times(20-0) $
$=10,000\, cal$
Heat required to raise the temperature of ice upto $0^{\circ} C$
$\Delta Q_{2} =200 \times 0.5 \times 20 $
$=2000 \, cal $
$\Delta Q_{1} -\Delta Q_{2}=8000\, cal$
Melts the ice
$8000=m \times 80 $
$m=100\, g$
So, mass of water is $500+100=600\, g$.