P21 005

5. We use the following relation derived in Sample Problem 21-2:

Tf
"S = mc ln .
Ti
(a) The energy absorbed as heat is given by Eq. 19-14. Using Table 19-3, we find

J
Q = cm"T = 386 (2.00 kg)(75 K) = 5.79 × 104 J
kg·K
where we have used the fact that a change in Kelvin temperature is equivalent to a chance in Celsius
degrees.
(b) With Tf = 373.15 K and Ti = 298.15 K, we obtain

J 373.15
"S =(2.00 kg) 386 ln = 173 J/K .
kg·K 298.15

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