A calorimeter contains 400gm of water at a temperature of $5 ° C$ . The 200gm of water at a temperature of $+ 10 ° C$ and 400gm of ice at a temperature of $- 60 ° C$ are added. Specific heat capacity of water = 1000 cal/kg K. Specific latent heat of fusion of ice $80 \times 1000$ cal/kg. Relative specific heat of ice = 0.5. Ratio between masses of ice and water in equilibrium state is ______

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Detailed Solution

Amount of heat released

Water : $Q1=400×1×5=2000 cal$

$Q2=200×1×10=2000 cal$

Amount of required

Ice : $Q3=400×0.5×60=12000 cal$

$(Q1+Q2)$ released < Q3 required

Difference in Q = 8000 cal

$8000=m \times Lf$

$8000=m \times 80$

M = 100 g

100 g of water changes into ice at $0^{0} C$

Final contents : $600g-100g=500g$ water

$400g+100g=500g ice$

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