Answer: E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
Explanation:
The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH
E = W/QH.
W=QH – QC,
Where Qc is the output heat.
That is,
E=1 - Qc/QH
E =1 - Tc/TH
where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.
Note: The unit of temperature must be in Kelvin.
Tc = 300K
TH = 2000K
Substituting the values of E, we have;
E = 1 - 300K/2000K
E = 1 - 0.15
E = 0.85
Therefore the efficiency is: E = 0.85 or 85%
Answer:
Vf = 0.0108 m³
Explanation:
Assuming there is no change in internal energy, we can calculate the changed volume by the following formula:
where,
W = Work = Energy Supplied = 1000 J
P = Pressure = 101325 Pa
Vf = Final Volume = ?
Vi = Initial Volume = 0.001 m³
Therefore,
<u>Vf = 0.0108 m³</u>
First one is B, second one is A
Answer:
The electrical energy is 13.5 kWh or 48600 kJ
Explanation:
Given that,
The power of the resistance heater, P = 4.5 kW
The water heater runs for 3 hours to raise the water temperature to the desired level.
We need to find the amount of electric energy used in both kWh and kJ.
Energy = Power × time
E = 4.5 kW × 3 h
= 13.5 kWh
Since,
1 kWh = 3600 kJ
13.5 kWh = 48600 kJ
Hence, the required electrical energy is 13.5 kWh or 48600 kJ