At 50 F,
,
which is very close to the distance between the towers. This is
because the sag is very small.
(a) From
,
.
(b) At
, the
slop of the cable is
. Therefore, the increase in
is approximately
4-4
Molecular weight of water is , so
is
. From ,
.
4-5
From
,
, which means that
.
4-6
From , we find that the mole number of the gas is
. (a) With the given initial pressure, volume, and
temperature
, the spring constant , and the cross-sectional area ,
we set up an equation of state from the new pressure and the new
volume formed by the increase in the height :
where .
Solving this quadratic equation for , we get
(b) The new pressure is
.
5-1
Let the specific heat of the metal , and that of water .
At the end of the day (at equilibrium), the heat gain of the
water (
) + container (
) is the same
as the heat loss of the metal (). The initial
temperature of the water+container is and the final
temperature
. The initial temperature of the small
metal was
.
If you solve this equation for , you get
.
5-2
The intensity of the sunlight:
.
Mass of the water: . Specific heat of the water:
. Temperature increase:
.
Duration of exposure:
. Efficiency:
. The unknown area of the collector: .
Then we can set up an equation of the total received energy and the
heat gain by the water:
So
5-3
Latent heat of the water:
(vapourisation) and
. Specific heat
of water:
. Mass of the ice:
. The unknown mass of the steam: . The temperature
difference for both:
.
The gain and the loss of heat by the ice and the steam are equal:
So
.
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Hyok-Jon Kwon
2001-08-05