AP EAPCET 23-Aug-2021 Shift 1 Paper

Given, mass of steam, mλ=5g
Temperature of steam, Ts=100°C
Mass of ice, mi=5g
Temperature of ice, Ti=0°C
Since, latent heat of vaporisation, lv=540‌cal∕g
Latent heat of fusion, lj=80‌cal∕g
Now, heat given by steam during fully conversion into water, at 100°C
and Q1=mj×540=2700‌cal
Heat taken by ice to get fully converted into water at 100°C.
Q2=milf+mi×s×ΔT
=5×80+5×1×100 (∵s=1‌for‌water)
=400+500=900
As, Q2<Q4
∴ Temperature of mixture will be 100°C.