NEET Class neet Answered
Please answer the following question with explanation
Asked by deepakudgiri29 | 02 May, 2019, 11:08: PM
Expert Answer
Change in entropy ΔS = entropy change for ice to melt + entropy change when heated from 0°C to 40°C
= (ΔQ/Ti) + m×Cp×ln( Tf / Ti)
ΔQ = latent heat of fusion of ice = 80 cal/g
Ti = ice temperature = 273 K , Tf = final temperature = 313 K
m = mass of ice = 1 g
CP = Specific heat of ice = 1 cal/( g °C )
ΔS = (80/273) + 1×1×0.1366 ≈ 0.42 cal/°C
Answered by Thiyagarajan K | 03 May, 2019, 02:22: PM
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