A piece of ice of mass 40g is added to 200g of water at 50 degree celcius

Asked by surendrakumarclw | 20th Mar, 2021, 06:16: PM

Expert Answer:

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 | 20th Mar, 2021, 07:16: PM