NEET 2019 Model Question Paper 1

Here, Specific heat of water,
sw=1calg−1°C−1
Latent heat of steam, Ls=540calg−1
Heat lost by m g of steam at 100ºC to change into water at 80ºC is
Q1=mLs+mswΔTw
=m×540+m×1×(100−80)
=540m+20m=560m
Heat gained by 20 g of water to change its temperature from 10ºC to 80ºC is
Q2=msswΔTw=20×1×(80−10)=1400
According to principle of calorimetry,
Q1=Q2
Therefore, 560 m = 1400 or m = 2.5 g
Total mass of water present = (20+m) g = (20 + 2.5) g = 22.5g