How much energy is required to change a 39.0-g ice cube from ice at - 14.0°c to steam at 108°c?

Question

Mariam Spence

Physics

31 December, 06:29

How much energy is required to change a 39.0-g ice cube from ice at - 14.0°c to steam at 108°c?

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Answers (1)

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Nick Bowman · Answer 1

Heat is added to the mass as follows:

Q_-14-0 = mC_iΔT = 39*2.06 * (0--14) = 39*2.06*14 = 1124.76 J

Q_0 = mC_f = 39*334 = 13026 J

Q_0-100 = mC_wΔT = 39*4.18*100 = 16302 J

Q_100 = mC_v = 39*2230 = 86970 J

Q_100-108 = mC_sΔT = 39*2.03 * (108-100) = 39*2.03*8 = 633.36J

Q = Summation of all the heats added = 1124.76+13026+16302+86970+633.36 = 118056.12 J ≈ 118.06 kJ