5. We use the following relation derived in Sample Problem 212:
∆
S
=
mc
ln
µ
T
f
T
i
¶
.
(a) The energy absorbed as heat is given by Eq. 1914. Using Table 193, 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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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy, Heat

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