Physics 200
Practice Final
(Sketchy, as in outlined) Solution
I.
(i) In time
t
the heater will have provided energy
Q
heater
=
Pt
, where
P
is the power. We
frst compare this to the amount o± energy needed to get the water to the boiling point (the
symbols should be relativity sel±explanatory; we will keep all the quantities positive here):
Q
boil
= (
m
Al
c
Al
+
m
water
c
water
)Δ
T
boil
,
which turns out to be larger than
Q
heater
. So the water will not boil, and we can fnd the
fnal temperature with
T
=
T
0
+ Δ
T
=
T
0
+
Q
heater
m
Al
c
Al
+
m
water
c
water
.
Let the the denominator be
α
±or the next part.
(ii) It’s clear this time that the water will not get to the boiling point. Let
R
be the rate o± rain,
so that the amount o± rain±all is
Rt
, and let
T
′
be the fnal temperature with the rain mixed
in. Then
α
(
T
′

T
) + (
Rt
)
c
water
(
T
′

T
rain
) = 0
,
which gives
T
′
=
αT
+
Rtc
water
T
rain
α
+
Rtc
water
.
II. Let
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 Fall '08
 RAMAMURTISHANKAR
 Physics, Energy, Kinetic Energy, Potential Energy, Power, Heat, final temperature

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