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3. We use the following relation derived in Sample Problem 202:
=l
n
.
§·
Δ
¨¸
©¹
f
i
T
Sm
c
T
(a) The energy absorbed as heat is given by Eq. 1914. Using Table 193, we find
()
(
)
4
J
=
= 386
2.00 kg 75 K = 5.79 10 J
kg K
Δ×
⋅
Qc
mT
where we have used the fact that a change in Kelvin temperature is equivalent to a change
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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
 Any
 Physics, Energy, Heat

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