Answer:
24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)
Explanation:
24) This problem can be solved by means of the following equation.
where:
DU = internal energy difference [J]
Q = Heat transfer [J]
W = work [J]
Since there are no temperature changes the internal energy change is equal to zero
DU = 0
therefore:
The work is equal to the heat transfered, W = 75 [J].
25) The heat transfer can be calculated by means of the following equation.
![Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]](https://tex.z-dn.net/?f=Q%20%3D%20m%2Ac_%7Bp%7D%2ADT%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%200.4%5Bkg%5D%5C%5Cc_%7Bp%7D%20%3D%20specific%20heat%20%3D%20897%5BJ%2Fkg%2AK%5D%5C%5CDT%3D%205%20%5BC%5D)
Q = 0.4*897*5 = 1794[J]
Work is equal to heat transfer, W = 1794[J]
26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.
where:
m = mass = 0.5[kg]
g = gravity = 9.81[m/s^2]
h = 1.5 [m]
![E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D0.5%2A9.81%2A1.5%5C%5CE_%7Bp%7D%3D7.36%5BJ%5D)
The heat developed can be calculated by means of the following equation.
![Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]](https://tex.z-dn.net/?f=Q%3Dm%2Ac_%7Bp%7D%2ADT%5C%5CQ%3D0.5%2A130%2A1%5C%5CQ%3D65%5BJ%5D)
The number of times will be calculated as follows
n = 65/7.36
n = 8.8 (times) or 9 (times)
D transferring electrons because that causes electricity
Answer:
Power output: W=1426.9MW
Explanation:
The power output of the falls is given mainly by its change in potential energy:
The potential energy for any point can be calculated as:
If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:
If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:
Complete Question
A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.
What is the refrigerator's coefficient of performance? COP
(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP
Answer:
a
b
Explanation:
From the question we are told that
The lower operation temperature of refrigerator is 
The upper operation temperature of the refrigerator is 
Generally the refrigerators coefficient of performance is mathematically represented as
=> 
=>
Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as
=> 
=>
<span>It stores energy and delivers it in a short burst.
The whirring sound is produced by the charging of the capacitor. A capacitor is an electrical component which is capable of storing charge. When the capacitor stores charge, it is storing energy. After doing so, the capacitor releases the electrical energy that it had stored as light energy, which is seen as the flash of the camera. It must do so in a burst, because the intensity of the flash is very high and would require a high amount of energy to maintain.
</span>
