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$2\, kg$ of ice at $-20°C$ is mixed with $5\, kg$ of water at $20°C$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are $1\, kcal/kg\, per °C$ and $0.5\, kcal/kg/°C$ while the latent heat of fusion of ice is $80\, k\,cal/kg$ ........ $kg$
$7$
$6$
$4$
$2$
Solution
(b) Initially ice will absorb heat to raise it's temperature to $0°C$ then it's melting takes place
If $m_i =$ Initial mass of ice, $m_i' =$ Mass of ice that melts and $m_W =$ Initial mass of water
By Law of mixture Heat gained by ice $=$ Heat lost by water ==> ${m_i} \times c \times (20) + {m_i}'\, \times L$ = ${m_W}{c_W}[20]$
==> $2 \times 0.5(20) + {m_i}' \times 80$ = $5 \times 1 \times 20$ ==> ${m_i}'$$= 1\,kg$
So final mass of water = Initial mass of water $+$ Mass of ice that melts $= 5 + 1= 6\, kg.$
