Question

Let’s say you try to get crafty. Instead of directly heating your bromoform with the Bunsen Burner, you decide it would be better to drop in a hot chunk of Europium (d = 5.264 g/mL , c = 0.182 J/g*K ) metal. If you have a 47.37 g chunk of euroium, how hot (in K) must you heat the europium in order for the chunk of metal to have enough energy to heat a 4.24 mL sample of bromoform at room temperature (298 K) to it's boiling temperature?

The boiling point of bromoform is 422.20 K.
Bromoform density = 2.89 g/cm³  specific heat = 0.516 J/g*K

Answer 1

Temperature to which Europium must be heated = 389.1 K

Explanation

Volume of bromoform = 4.24 mL = 4.24 cm³

density of bromoform = 2.89 g/cm³

mass of bromoform = (volume of bromoform) * (density of bromoform)

mass of bromoform = (4.24 cm³) * (2.89 g/cm³)

mass of bromoform = 12.2536 g

Energy required to heat bromoform = (mass of bromoform) * (specific heat of bromoform) * (final temp. - initial temp.)

Energy required to heat bromoform = (12.2536 g) * (0.516 J/g.K) * (422.20 K - 298 K)

Energy required to heat bromoform = 785.3 J

Energy transferred by Europium = Energy required to heat bromoform

Energy transferred by Europium = 785.3 J

Temperature change of Europium = (Energy transferred by Europium) / [(mass Europium) * (specific heat Europium)]

Temperature change of Europium = (785.3 J) / [(47.37 g) * (0.182 J/g.K)]

Temperature change of Europium = 91.1 K

Final temperature of Europium = (Temperature change of Europium) + (initial temperature of Europium)

Final temperature of Europium = (91.1 K) + (298 K)

Final temperature of Europium = 389.1 K