The kinetic energy is transferred to thermal energy through friction
Answer:
- maintain its state of motion
- Keep its velocity constant (either at zero or non-zero value)
use the formula: v^2=(3kT)/m
Where:
<em>v is the velocity of a molecule</em>
<em>k is the Boltzmann constant (1.38064852e-23 J/K)</em>
<em>T is the temperature of the molecule in the air</em>
<em>m is the mass of the molecule</em>
For an H2 molecule at 20.0°C (293 K):
v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)
v^2 = 3.65e+6 m^2/s^2
v = 1.91e+3 m/s
For an O2 molecule at same temp.:
v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)
v^2 = 2.28e+5 m^2/s^2
v = 478 m/s
Therefore, the ratio of H2:O2 velocities is:
1.91e+3 / 478 = 4.00
Answer:
Explanation:
Given that
Temperature of warm water ,T₁ = 25 °C
Temperature of cold water ,T₂ = 5 °C
The maximum possible efficiency of any heat is will be same as the efficiency of the Carnot heat engine.Therefore the maximum efficiency of OTEC system is given as follows
T should be Kelvin scale
Now by putting the values
Therefore the maximum efficiency will be 0.067 or 6.7%.
