What is the temperature change of the skillet?

Question

Exactly 18.4 mL of water at 31.0 °C is 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?

Answer 1

According to the law of conservation of energy, the heat LOST by the hot skillet must equal the heat gained by the colder water. So, we begin this problem by finding out how much heat is gained by the water. The equations we will use for this are q = mC∆T and q = m∆Hvap

Step 1: heat needed to raise temperature of 18.4 ml water from 31.0º to 100º:

q = mC∆T

q = heat = ?

m = mass of water = 18.4 ml x 1 g / ml = 18.4 g (assuming a density of 1g/ml for water)

C = specific heat of water = 4.184 J/gº (tabular value - look it up)

∆T = change in temperature of water = 69º

q = (18.4 g)(4.184 J/gº)(69º)

q = 5312 J

Step 2: heat needed to convert water to steam at 100º:

q = m∆Hvap

q = ?

m = mass = 18.4 g

∆Hvap = heat of vaporization for water = 2260 J/g (tabular value - look it up)

q = (18.4 g)(2260 J/g)

q = 41584 J

Step 3: Sum the heats required:

TOTAL HEAT = 5312 J + 41584 J = 46,896 J

We now set this to the heat that must be lost by the hot skillet, and the heat lost by the skillet is

q = mC∆T for the skillet, but note that we are given mass in kg and C in J/molº

Let us convert so we have units that are consistent with one another:

25.19 J/molº x 1 mol iron/55.85 g 0.451 J/gº (now in units using g which is same as mass of iron)

q = heat = 46,896 J (see above: this is the heat gained by the water)

m = mass of iron skillet = 1.10 kg x 1000 g / kg = 1100 g

C = specific heat of iron skillet = 0.451 J/gº (see conversion above)

∆T = change in temperature of the iron skillet = ?

46,896 J = (1100g)(0.451 J/gº)(∆T)

∆T = 94.5º = temperature change of the skillet