a given sample of a gas has a volume of 3.0 L at a pressure of 4.0 atm. If the temperature remains constant and the pressure is
1 answer:
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.
<u>Explanation:</u>
We are given:
Rate of flow of ideal gas , n = 4 kmol/hr = (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)
Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)
We know that:
(For isothermal process)
So, by applying first law of thermodynamics:
.......(1)
Now, calculating the work done for isothermal process, we use the equation:
where,
= change in work done
n = number of moles = 1.11 mol/s
R = Gas constant = 8.314 J/mol.K
T = temperature = 475 K
= initial pressure = 100 kPa
= final pressure = 50 kPa
Putting values in above equation, we get:
Calculating the heat flow, we use equation 1, we get:
[ex]\Delta q=3.038kW[/tex]
Now, calculating the rate of lost work, we use the equation:
Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.
Answer:
The percentage efficiency of the electrical element is approximately 82.186%
Explanation:
The given parameters are;
The thermal energy provided by the stove element, = 3.34 × 10³ J
The amount thermal energy gained by the kettle, = 5.95 × 10² J
The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;
Therefore, we get;
The percentage efficiency of the electrical element, η% ≈ 82.186%.
Answer:
<h3>The answer is 2 g/cm³</h3>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 48 g
volume = 24 cm³
We have
We have the final answer as
<h3>2 g/cm³</h3>Hope this helps you