Question

Exactly 10.3 mL of water at 31.0 degree C are added to a hot iron skillet. All of the water is converted into steam at 100 degree C. The mass of the pan is 1.15 kg and the molar heat capacity of iron is 25.19 J/(mol middot degree C). What is the temperature change of the skillet? The temperature of the skillet goes down, not up. Remember, the heat gained by the water is LOST by the skillet: Q_skillet = -Q_water

Answer 1

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