Question

Exactly 16.7 mL of water at 32.0°C are added to a hot iron skillet. All of the water is converted into steam at 100°C. The mass of the pan is 1.10kg and the molar heat capacity of iron is 25.19J/(Mol x °C). What is the temperature change of the skillet?

Answer 1

The heat gained by water = heat lost by skillet

mcdt + mL = m'c'dt'

Where

m = mass of water = density x volume of water

= 1.0 g/ml x 16.7 mL = 16.7 g

c = specific heat capacity of water = 4.186 J/goC

dt = change in temperature of water = 100 - 32 = 68 oC

L = Heat of Vaporization of water = 2260 J/g

m' = mass of skillet = 1.10 kg = 1100 g

c' = molar heat capacity of iron =25.19J/(Mol x °C) = 25.19 /55.8 J/goC

= 0.451 J / goC

dt' = change in temperature of iron skillit = ?

Plug the values we get dt' = 38.6 oC