Question

# Exactly 17.7 mL of water at 35.0 °C are added to a hot iron skillet. All...

Exactly 17.7 mL of water at 35.0 °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 1.20 kg and the molar heat capacity of iron is 25.19 J/(mol·°C). What is the temperature change of the skillet?

The heat balance:

Qgain = -Qlost

Qgain = water

Qlost = skillet

We need two types of heat for water, since it is vporized

latent heat = heat required to boil water

sensible heat, heat required to change temperature of water

then

Qskillet = Qwater

Qskillet = Qsensible + Qlatent

mskillet * Cp skillet * (dT) = mwater* Cpwater*(Tf-Twater) + mwater * Latent heat of vapor

LH water = 2264.76 J/g.

substitute data

mskillet * Cp skillet * (dT) = mwater* Cpwater*(Tf-Twater) + mwater * Latent heat of vapor

change molar heat capacity to mass capacity

25.19 J/mol K * 1 mol o iron / 55.85 g = 0.451 J/gC

mskillet * Cp skillet * (dT) = mwater* Cpwater*(Tf-Twater) + mwater * Latent heat of vapor

1200* 0.451 * (dT) = 17.7* 4.184*(100-35) + 17.7*2264.76

dT = (17.7* 4.184*(100-35) + 17.7*2264.76) / (1200* 0.451)

dT = 82.96 is the change in T for skillet