Answers
under adiabatic conditions, heat gained by water= -heat lost by iron pan
heat gained by water= from 31 deg.c to 100 deg.c ( sensible heat)+ latent heat of vaporization.
mass of pan = 1.15 kg= 1.15*1000 gm, atomic weight of iron ( pan material)= 56
moles of iron pan= 1.15*1000/56 =20.54 moles
specific heat of pan = 25.19 J/mole.deg.c. Let the temperature change= deltaT
hence heat lost by pan = 20.54*25.19*deltaT
Density of water= 1 gm/ml, assumed. Mass of water= Volume*density= 10.3*1gm/ml= 10.3 gm, moles of water= mass/molar mass= 10.3/18= 0.572, specific heat of liquid water= 4.184 J/gm.deg.c, latent heat of water= 40.65 Kj/mole
Heat gained by water : 1. Sensible heat= mass of water* specific heat of water* temperature difference (100-31) = 10.3*4.184*(100-31)= 2974 joules
2. Latent heat of water= moles of water* latent heat of vaporization = 40.65*0.572 KJ =23.25 Kj= 23.25*1000Joules
total heat transferred to water= 2974+23.25*1000 =26225.37 J
hence 20.54*25.19*deltaT`=-26225.37
deltaT= 26225.37/(20.54*25.19)= -50.67 deg.c
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