ch18-p034

# Ch18-p034 - | Q | = P t =(900 J/min(30 min = 27000 J = L m Thus with m = 0.40 kg we find L = 67500 J/kg ≈ 68 kJ/kg(b During the final 20 minutes

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34. While the sample is in its liquid phase, its temperature change (in absolute values) is | Δ T | = 30 °C. Thus, with m = 0.40 kg, the absolute value of Eq. 18-14 leads to | Q | = cm | Δ T | = ( 3000 J/ kg C ⋅° )(0.40 kg)(30 C ° ) = 36000 J . The rate (which is constant) is P = | Q | / t = (36000 J)/(40 min) = 900 J/min, which is equivalent to 15 Watts. (a) During the next 30 minutes, a phase change occurs which is described by Eq. 18-16:
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Unformatted text preview: | Q | = P t = (900 J/min)(30 min) = 27000 J = L m . Thus, with m = 0.40 kg, we find L = 67500 J/kg ≈ 68 kJ/kg. (b) During the final 20 minutes, the sample is solid and undergoes a temperature change (in absolute values) of | Δ T | = 20 C°. Now, the absolute value of Eq. 18-14 leads to c = |Q| m | Δ T| = P t m | Δ T| = (900)(20) (0.40)(20) = 2250 J kg·C° ≈ 2.3 kJ kg·C° ....
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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