Exactly 18.4 mL of water at 31.0 degree C are added to a hot iron skillet. All o

Question

Exactly 18.4 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 1.50 kg and the molar heat capacity of iron is 25.19 J/(mol middot degree C). What is the temperature change of the skillet? You actually don't need to know the initial temperature of the iron skillet since you are only asked for Delta T, not the actual final T. All you need to do is calculate the total amount of heat needed to BOTH raise the temperature of the water to 100 degree C and then boil it. This total amount of heat came FROM the skillet and must cause the temperature of the skillet to decrease.

Explanation / Answer

heat: q = (18.4 g) (100-31 °C) (4.184 J/g °C) = 5312.00J = 5.312kJ

heat of vaporaisation of water = 40.7 kj/mol

Hvap = 40.7 kJ/mol
The molar mass of H2O = 18.0 gram/mol

boil: q = (40.7 kJ/mol) (18.0 g / 18.0 g/mol) = 40.7 kJ

total: 40.7 kJ + 5.312kJ = 46.012 kJ

46122.464 J = (26.86mol) (t) (25.19 J/mol °C)

t = 68.16°C

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.