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10 gm of ice at -10℃ is added to 10gm of water at 85℃.What is the final temperature?

Asked by akpdbbhhs | 14 May, 2021, 12:51: PM
Let us assume all ice will melt and we get final equilibrium temperature T oC

Heat gain by ice = mice × ( Cp_ice × 10 + L ) + mice × Cp_w  × T

where mice is mass of ice , Cp_ice = 2100 J/ kg is specific heat of ice,  L = 336 kJ/ kg is latent heat of fusion of ice
and Cp_w = 4200 J/kg is specific heat of water

Heat gain by ice  = 0.01 × ( 2100 × 10  + 336 × 103 ) + 0.01 × 4200  × T = 3570 + 42T

Heat loss by water = mwater × Cp_w × (85 - T ) = 0.01 × 4200 × (85-T) = 3570 - 42T

If we equate heat gain by ice to heat loss by water , we get T = 0 oC

Answered by Thiyagarajan K | 16 May, 2021, 05:47: PM

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