$2 k g$ of ice at $- 20^{\circ} C$ is mixed with $5 k g$ of water at $20^{\circ} C$ in an insulating vessel having a negligible heat capacity. Calculate final mass (in $k g$ ) of water remaining in the container. It is given that the specific heat of water and ice are $1 k c a l / k g^{\circ} C$ and $0.5 k c a l / k g^{\circ} C$ , while the latent heat of fusion of ice is $80 k c a l / k g$ .

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Solution

We know that: $Q = m s Δ θ$

Given: $m_{1} = 2 k g, θ_{1} = - 20^{\circ} C, m_{2} = 5 k g, θ_{2} = 20^{\circ} C, s_{2} = 1 k c a l / k g^{\circ} C, s_{1} = 0.5 k c a l / k g^{\circ} C, L = 80 k c a l / k g$

Maximum heat that can be released by water

$m_{2} s_{2} (20 - 0) = 5 \times 1 \times (20 - 0) = 100 k c a l \dots (i)$

Heat required to change ice from $- 20^{\circ} C$ to $0^{\circ} C$

$m_{1} s_{1} (0 - (- 20)) = 2 \times 0.5 [0 - (- 20)] = 20 k c a l \dots (i i)$

We know that: $Q = m L$

Heat required to melt ice $= 2 \times 80 = 160 k c a l \dots (i i i)$

From $(i)$ , $(i i)$ and $(i i i)$

$100 k c a l < (20 + 160) k c a l$ ,

Whole ice is not melted.

Equilibrium temperature is $0^{\circ} C$

Amount of ice melted
$= \frac{100 - 20}{80} = 1 k g$

Mass of water at equilibrium temperature will be $5 + 1 = 6 k g$ .

Final answer: $6 k g$ .

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2 kg of ice at $- 20^{o} C$ is mixed with 5 kg of water at $20^{o} C$ 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 $k c a l / k g /^{o} C$ and 0.5 $k c a l / k g /^{o} C$ while the latent heat of fusion of ice is 80 kcal/kg