10 gm of ice at -10℃ is added to 10gm of water at 85℃.What is the final temperature?
Asked by akpdbbhhs | 14th 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 | 16th May, 2021, 05:47: PM
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