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Question

A lump of 0.10kg of ice at 10C is put in 0.15kg of water at 20C. How much water and ice will be found in the mixture when it has reached in thermal equilibrium.
(Specific heat of water =1kcalkg1, specific heat of ice =0.50kcalkg1 while its latent heat =80kcalkg1)

A
Ice is 68.75gm, and water is 181.25gm at 0C.
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B
Ice is 16.75gm, and water is 11.25gm at 0C.
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C
Ice is 108.75gm, and water is 101.25gm at 10C.
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D
Ice is 6.75gm, and water is 181.25gm at 100C.
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Solution

The correct option is A Ice is 68.75gm, and water is 181.25gm at 0C.
The heat, which 0.15kg of water can release when its temperature is changed from 20C to 0C.
Q1=mWsWΔTW
where, mW is the mass of the water, sW is the specific heat of water and ΔTW is the change in temperature of water.
Given, mW=0.15 kg, sW=1 kcal kg1.
Q1=0.15×(1×103)×(200)
=3000cal ...(1)
Now, heat absorbed by 0.10kg of ice at 10C to increase its temperature to 0C.

Q2=miSiΔTi
where mi is the mass of the ice, si is the specific heat of ice and ΔTi is the change in temperature of ice.
Given, mi=0.10 kg, si=0.50 kcal kg1
Q2=0.10×(0.5×103)×[0(10)]
=500cal
So, remaining heat,
Q=Q1Q2=3000500=2500 cal

Now as latent heat of ice is l=80kcalkg1, the remaining heat will melt ice only
L=Qmm=Qm=250080=31.25 g of ice.
Initial amount of ice =0.10 kg=100 g
So, the remaining ice,
=10031.25=68.75g
Initial amount of water =0.15 kg=150 g
Therefore, total water =150+31.25=181.25g
The temperature of the system will be 0C.

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