Post by: ed_saifa on April 29, 2009, 10:25:03 am
Space invaders land on earth. On the invaders' temperature scale, the ice point is at 24.0 °I (I = invader), and the steam point is at 156 °I. The invaders' thermometer shows the temperature on earth to be 66.6 °I. What would this temperature be on the Celsius scale?
Ans:32.3
When you take a bath, how many kilograms of hot water (40.8 °C) must you mix with cold water (11.0 °C) so that the temperature of the bath is 31.5 °C? The total mass of water (hot plus cold) is 192 kg. Ignore any heat flow between the water and its external surroundings.
Ans:132kg
A woman finds the front windshield of her car covered with ice at -11.5 °C. The ice has a thickness of 2.18 x 10-4 m, and the windshield has an area of 1.32 m2. The density of ice is 917 kg/m3. How much heat in joules is required to melt the ice?
Ans:94500
A 13.3-kg block of ice has a temperature of -20.5 °C. The pressure is one atmosphere. The block absorbs 6.76 x 106 J of heat. What is the final temperature in degrees Celsius of the liquid water?
Ans: 31.6
Post by: enwiabe on April 29, 2009, 12:09:32 pm
When C = 0, I = 24, and when C = 100, I = 156.
therefore, 24a + b = 0 (1)
and 156a + b = 100 (2)
Taking eqn 1,
24a = -b
it turns out that 156/24 = 6.5
so 6.5 * 24a = 6.5 * - b
156a = -6.5b
Going back to equation 2:
156a + b = 100
-6.5b + b = 100
-5.5b = 100
therefore b = -100/5.5 = -200/11 ~ -18.18
Subbing back into 24a = -b
a = -b/24 = -(-200/11)/24 = 25/33 ~ 0.76
So now we have our relationship!
C = 25*I/33 - 200/11
so when earth's temperature = 66.6 degrees = 333/5 degrees
333/5 = 25I/33 - 200/11
25I/33 = 333/5 + 200/11
= (3663 + 1000)/55
= 4663/55
I = 33*4663/(25*55)
= 3*4663/(25*11)
= 13989/275
~ 50.87 degrees Invader
Post by: ed_saifa on April 29, 2009, 02:48:13 pm
Post by: ed_saifa on May 09, 2009, 01:18:23 pm
Light in a vacuum is incident on a transparent glass slab. The angle of
incidence is 35.0°. The slab is then immersed in a pool of liquid. When the
angle of incidence for the light striking the slab is 20.3°, the angle of
refraction for the light entering the slab is the same as when the slab was in a
vacuum. What is the index of refraction of the liquid?
Ans: 1.65
A point source of light is submerged 2.2 m below the surface of a lake and
emits rays in all directions. On the surface of the lake, directly above the
source, the area illuminated is a circle. What is the maximum radius that this
circle could have?
Ans: 2.5m
Post by: Mao on May 09, 2009, 07:20:40 pm
for slab in vacuum:
for slab in liquid:
2.
finding the critical angle:
hence, the radius is
Post by: ed_saifa on May 27, 2009, 10:30:26 am
Ans: 8.515726717169m
Cheers