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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.
- Spring '08