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Standard XII

Physics

Latent Heat

2 kg of ice a...

Question

$2$ kg of ice at $- 20^{o}$ C is mixed with $5$ kg of water at $20^{o}$ C. The water content of the final mixture is (Latent heat of fusion $= 80$ kcal/kg, specific heat of water $= 1$ kcal/kg- $^{o} C$ , specific heat of ice $= 0.5$ kcal/kg- $^{o} C$ .

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Solution

The correct option is B $6$ kg
Formulae:

$h e a t = m s Δ t$

m $\times$ latent heat = heat remained

Heat remained = Heat lost - Heat gained

$= (5 \times 1 \times 20) - (2 \times 0.5 \times 20)$

$= 100 - 20 = 80 k c a l$

$m \times 80 = 80$

$\Rightarrow m = 1 k g$

Therefore final mass of water $= 5 + 1 = 6 k g$

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2 kg of ice at −20oC is mixed with 5 kg of water at 20oC. The water content of the final mixture is (Latent heat of fusion =80 kcal/kg, specific heat of water =1 kcal/kg-oC, specific heat of ice =0.5 kcal/kg- oC.

The correct option is B 6 kgFormulae:heat=msΔtm× latent heat = heat remainedHeat remained = Heat lost - Heat gained=(5×1×20)−(2×0.5×20)=100−20=80kcalm×80=80⇒m=1kgTherefore final mass of water =5+1=6kg

$2$ kg of ice at $- 20^{o}$ C is mixed with $5$ kg of water at $20^{o}$ C. The water content of the final mixture is (Latent heat of fusion $= 80$ kcal/kg, specific heat of water $= 1$ kcal/kg- $^{o} C$ , specific heat of ice $= 0.5$ kcal/kg- $^{o} C$ .