A drop of water falls from the top of a water fall that is 50 meters high above a lake.
Assume the gravitational acceleration g = 10 m s^{-2},
and the air resistence to be zero. Further more, assume that the water
drop does not evaporate any watermolecule during its decent.
What is the speed of the water drop just before it hits the lake surface?
m g h = m v^{2} / 2
Thus,
v = Ö (2 g h)
= = Ö (2 * 10 * 50)
= 31.6 m s^{-1}
Derive the units please.
What is the increase in temperature if all the kinetic energy of the
water drop is used to heat up the water drop?
Assume one kg of water (or any amount)
Energy = m g h J
IF the temperature increase is dT, then
Energy = m dT * 4184
dT = g h / 4184 (Independent of mass)
= 0.12 K or C (a measurable difference)
Assume the efficiency of converting hydro-energy to electric energy to be 100%.
How much water from the fall (of 50 m height) is required to be diverted
in order to generate one-kilowatt-hour of electric energy?
1 kwh = 1000 * 3600 J
= 3.6e6 J
= m g h Thus,
m = 3.6e6 J / (10 m s^{-2} * 50 m)
= 7200 kg
Assume the efficiency of converting hydro-energy to electric energy to be 50%.
How much water from the fall (of 50 m height) is required to be diverted in order to generate one kilowatt-hour of electric energy?
Since the efficiency is 0.5, twice the amount of the above is required.
7200 / 0.5 = 14400 kg
When 2.00 grams of H_{2} is oxidized, 242 kJ of energy is released.
How much water is produced in order to generate one kilowatt-hour (3.6e6 J)
of energy?
When 2.0 g of H_{2} is oxidized, 18.0 g of water is generated.
3.6e6 J * 18.0 g H_{2}O / 2.42e5 J
= 268 g H_{2}O produced
The efficiency of converting chemical energy into electric energy is 30%.
How much hydrogen gas is required in order to generate one kilowatt-hour of energy?
The amount of H_{2} required
= 3.6e6 J * 2.0 g H_{2} / 2.42e5 J / 0.3
= 99.3 g H_{2} required
A photon in the X-ray region has a wavelength of 1.0e-10 m.
What is the frequency?
Understand this formula, and you'll remember it
Frequency = speed of light / wavelength = 3e8 m s^{-1} / 1.0e-10 m
= 3e18 s^{-1} (or Hz)
What is the energy of this photon?
Energy of photon = h * frequency = 6.62619e-34 J s * 3e18 s^{-1} = 1.99e-15 J (look small, but this energy is for 1 photon.)
A particle with a mass of 1.67e-30 g has a kinetic energy of the photon, what is the velocity of this particle?
Kinetic energy = 0.5 m * v^{2} = 1.99e-15 J
v^{2} = 1.99e-15 J / (0.5 * 1.67e-33 kg)
= 2.38e12
v = 1.54e9 m s^{-1}
The velocity is greater than the speed of light. We need another method
to calculate the velocity. We will talked about that method in a future
module. If you have pointed out this, that is great.
If all the energy of this photon is converted to two particles of equal mass, what is the rest mass of each particle?
The conversion factor is enclosed in ( ) below
1.99e-15 J * (1 amu / 1.4924e-10 J)
= 1.33e-5 amu
Half of the above (6.67e-6 amu) is the rest mass of one of the two particles created.
The value also depends on units used. Another answer is 1.11e-32 kg.