5. We use the following relation derived in Sample Problem 21-2: ∆ S = mc ln µ T f T i ¶ . (a) The energy absorbed as heat is given by Eq. 19-14. Using Table 19-3, we Fnd Q = cm ∆ T = µ 386 J kg · K ¶ (2 . 00 kg)(75 K) = 5 . 79 × 10 4 J where we have used the fact that a change in Kelvin temperature is equivalent to a chance in Celsius
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