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A piece of ice of mass 40g is added to 200g of water at 50 degree celcius
Asked by surendrakumarclw | 20 Mar, 2021, 06:16: PM
Let us assume ice is initially at 0o C

Heat gain by Ice = mice × (  L + Cp T ).....................(1)

where mice = 0.040 kg  is mass of ice , L = 334 kJ/kg  is latent heat of fusion of ice ,

Cp = 4186 J/(kg oC) is specific heat of water and T is final equilibrium temperature.

Heat loss by water = mw × Cp × ( 50 - T ).....................(2)

where mw = 0.2 kg is mass of water

By conservation of energy, Heat gain by ice = Heat loss by water

mice × (  L + Cp T ) = mw × Cp × ( 50 - T )

0.04 × ( 334 × 103 + 4200 T ) = 0.2 × 4200 × ( 50 - T )....................(3)

By solving above eqn.(3) , we get equilibrium temperature T = 28.4 oC
Answered by Thiyagarajan K | 20 Mar, 2021, 07:16: PM

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