Question

Exactly 14.1 mL of water at 31.0 degrees C are added to a hot iron skillet....

Exactly 14.1 mL of water at 31.0 degrees C are added to a hot iron skillet. All of the water is converted into steam at 100 degrees C. The mass of the pan is 1.45 kg and the molar heat capacity of iron os 25.19 J/(mol degrees C). What is the temperature change of the skillet?

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Answer #1

Heat lost by iron = heat gained by water

nFe X CFe X ΔT = energy needed to raise temperatute from 32oC to 100oC + heat needed vaporise the water.

nFe = no. of moles of iron, CFe = heat capacityof iron , ΔT = change in temperature.

nFe X CFe X ΔT = nH2O X CH2O X ΔT + ΔH X n

nFe = mass in grams / molar mass = 1450 / 55.84 = 25.96

nH2O =14.1 / 18 = 0.783 Moles 14mL of water = 14 g (because density = 1)

25.96 X 25.19 X ΔT = 0.783 X 75.24 X (100-31) + 40.65 X 0.783

ΔT = 4096.3 / 653.93 = 6.26oC

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