Answer:
The actual work output is 1,125 kilojoules.
Explanation:
Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.
Efficiency is calculated as:
efficiency = output / input
Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.
In this case:
- Efficiency always has a value between 0 and 1. In this case, efficiency=0.15
- output=?
- input= 7500 kilojoules
Replacing:
0.15=output/7500 kilojoules
Solving:
Output=0.15* 7500 kilojoules
output=1,125 kilojoules
<u><em>The actual work output is 1,125 kilojoules.</em></u>
The SI unit of energy is the Joule .
Any kind of energy ... electrical, mechanical, nuclear, solar,
wind etc. It's easy to change from one form to another ... we
do it every day ... and they all have the same unit.
Answer:
electrons pass through a cross section of the wire each second.
Explanation:
According to mole concept:
1 mole of an atom contains
number of particles.
Given : Charge on 1 electron =
Charge on 1 mole of electrons =
To calculate the charge passed we use the equation:
where,
I = current passed = 1 A
q = total charge = ?
t = time required = 1 sec
Putting values in above equation, we get:
When 96500C of electricity is passed , the electrons passed =
1 C of electricity is passed , the electrons passed =
Hence,
electrons pass through a cross section of the wire each second.
Answer:
(1) The maximum air temperature is 1383.002 K
(2) The rate of heat addition is 215.5 kW
Explanation:
T₁ = 17 + 273.15 = 290.15

T₂ = 290.15 × 3.17767 = 922.00139

Therefore,
T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K
The maximum air temperature = T₃ = 1383.002 K
(2)


Therefore;


Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg
Heat rejected per kilogram is given by the following relation;
= 0.718×(511.859 - 290.15) = 159.187 kJ/kg
The efficiency is given by the following relation;

Where:
β = Cut off ratio
Plugging in the values, we get;

Therefore;


Heat supplied = 
Therefore, heat supplied = 215491.064 W
Heat supplied ≈ 215.5 kW
The rate of heat addition = 215.5 kW.