Question

Exactly 18.2 mL of water at 23.0 °C are added to a hot iron skillet. All of the water is converted into steam at 100.0°C. The mass of the pan is 1.10 kg and the molar heat capacity of iron is 25.19 J/(mol·°C). What is the temperature change of the skillet? _____°C

Answer 1

lets calculate the heat required to convert water from 23 oC to steam at 100 oC

Ti = 23.0

Tf = 100.0

mass of water = density of water * volume of water

= 1.0 g/mL * 18.2 mL

= 18.2 g

here

Cl = 4.184 J/goC

Heat required to convert liquid from 23.0 to 100.0

Q1 = m*Cl*(Tf-Ti)

= 18.2 g * 4.18 J/goC *(100-23) oC

= 5863.46 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0

Q2 = m*Lv

= 18.2g *2260.0 J/g

= 41132 J

Total heat required = Q1 + Q2

= 5863.46J + 41132 J

= 46995 J

This much heat is required to convert water from 23 oC to steam at 100 oC

This heat must be supplied by hot iron

For iron:

m = 1.10 Kg = 1100 g

molar mass = 55.85 g/mol

mol of Fe = mass /molar mass

= 1100/55.85

= 19.7 mol

Q = n*c*delta t

46995 = 19.7 mol * 25.19 J/mol.oC*delta T

delta T = 94.7 oC

Answer: 94.7 oC