Answer:
w = 736 J
Explanation:
A Carnot cycle is a theoretical thermodynamic cycle that establishes the maximum efficiency of a heat engine operating between two heat sources at different temperatures.
The efficiency of a thermal engine is how much of the heat from the hot reservoir it can transform into work.
[tex]\eta = \frac{w}{q1}[/tex]
Where:
η: efficiency
w: work
q1: heat taken from the hot source
Therefore, the work can be calculated as:
[tex]w = \eta * q1[/tex]
We know the heat energy taken from the hot reservoir is the energy needed to raise the temperature of 1 kg of water in 1 K.
This is the definition of one Kcal.
q1 = 1 kcal
The efficiency of the Carnot cycle is determined by the temperatures of the reservoirs.
[tex]\eta = 1 - \frac{T2}{T1}[/tex]
These temperatures must be expressed in an absolute temperature scale (such as degrees Kelvin or degrees Rankine)
So in this case the work per cycle is:
[tex]w = q1 * (1-\frac{T2}{T1} )[/tex]
[tex]w = 1 kcal * (1-\frac{280 K}{340 K} ) = 0.176 kcal[/tex]
Now, work is not usually expressed in calories, so it should be converted to Joules:
1 kcal = 4184 J
[tex]w = 0.176 kcal * \frac{4184 J}{1 kcal} = 736 J[/tex]
The work done by the engine will be "736 J".
According to the question,
Mass, m = 1.0 kg
Temperature, T₁ = 280 K
T₂ = 340 K
Heat taken from source, q₁ = 1 Kcal
We know the relation, efficiency of thermal engine be:
→ η = [tex]\frac{w}{q_1}[/tex]
w = η × q₁
here,
η = 1 - [tex]\frac{T_2}{T_1}[/tex]
now,
→ w = q₁ × (1 - [tex]\frac{T_2}{T_1}[/tex])
By substituting the values, we get
= 1 × (1 - [tex]\frac{280}{340}[/tex])
= 0.176 kcal
Thus the above answer is appropriate.
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