This preview shows page 1. Sign up to view the full content.
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
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 SPRUNGER
 Physics, Energy, Heat

Click to edit the document details