UPSEE 2019 Solved Paper

Key-Idea The change in entropy of a system is given as ΔS=

ΔQ

T

where, T= temperature and ΔQ= heat absorbed by the system.
Here, latent heat of ice, L=80cal∕g, mass of water, m=10g and temperature,
T=0°C=273K
Heat given to 10g ice,
E=mL=10×80
⇒E=800cal
So, the entropy change
Δs=

E

T

=

800

273

⇒Δs=2.93cal∕K
Hence, the correct option is (b).