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Plzz answer this question of calorimetry 
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Asked by anjanakurup728 | 19 Oct, 2019, 19:22: PM
answered-by-expert Expert Answer
Specific heat of water Cpw = 4200 J/(kg °C)
 
Specific heat of ice Cpice = 2108 J/(kg °C)
 
Latent heat of ice, L = 334 kJ/kg
 
If one kg of ice at -10°C is melted completely by 1 kg of water at 100 °C and the final equilibrium temperature is T °C,
 
Heat gain by ice =  Cpice ×10 + L + Cpw T = 2018×10 + 334×103 + 4200 T ..............................(1)
 
Heat loss by water from 100°C to final equlibrium temperature T °C = Cpw ( 100 - T) = 4200 ( 100 - T ) .....................(2)
 
By equating (1) and (2), solving for T, we get T ≈ 7.7 °C
 
Mixture contenet at equilibrium, 2 kg of water at 7.7 °C
Answered by Thiyagarajan K | 20 Oct, 2019, 08:07: AM

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