Exactly 10.8 ml of water at 31.0 degree celcius are added to a hot iron skilllet....

Exactly 10.8 ml of water at 31.0 degree celcius are added to a hot iron skilllet. All of the water into converted into steam at 100 degree celcius. The mass of the pan is 1.00kg and the molar heat capacity is 25.19 J/mol degree celcius. What is the temperature change of the skillet?

Expert Solution

10.8 mL of water = 10.8 g because the density is given as 1 g/mL. 10.8 g = 10.8 g / 18 g/mole = 0.6 moles.

energy = 0.6 moles water * 74.3J/moleC * (100C - 23C) = 3432.6 J

Or, 74.3 J/mole C = 74.3 J/mole C / 18 g/mole = 4.128 J/gC (not typically the specific heat given for water but it's close).

energy = 10 g * 4.128 J/gC * (100C - 23C) = 3432.66 J, same answer.

Then, you need to add the energy to boil the water, that data isn't given in your question but it must have been provided. It will be the mass of water * heat of fusion or the moles of water * heat of fusion . Which one will depend on the units in the heat of fusion, J/g or J/mole.

The sum of those (the energy to boil the water into steam + 3432.66 J to heat the water to boiling) is the total energy given up by the skillet.

Again, you use energy = mass of skillet * specific heat of skillet * temperature change.

The energy is the total you just determined for the water. The skillet is 1.0 kg or 1000 g. Iron has an atomic mass of 55.8 g/mole so you have 17.92 moles Fe. Substitute and solve for the temperature change

total energy for the water = 17.92 moles Fe * 25.19 J/mole C * temperature change.
3432.66=17.92*25.19*delta T

temptarture change =7.60 ⁰C

queen_honey_blossom answered 2 years ago

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