Question

A heat engine operates on a Carnot cycle that runs clockwise between a reservoir at 340 K and a reservoir at 280 K. One cycle moves enough energy from the high-temperature reservoir to raise the temperature of 1.0 kg of water by 1.0 K. How much work is done by the engine in one cycle

Answer 1

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.

$\eta = \frac{w}{q1}$

Where:

η: efficiency

w: work

q1: heat taken from the hot source

Therefore, the work can be calculated as:

$w = \eta * q1$

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.

$\eta = 1 - \frac{T2}{T1}$

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:

$w = q1 * (1-\frac{T2}{T1} )$

$w = 1 kcal * (1-\frac{280 K}{340 K} ) = 0.176 kcal$

Now, work is not usually expressed in calories, so it should be converted to Joules:

1 kcal = 4184 J

$w = 0.176 kcal * \frac{4184 J}{1 kcal} = 736 J$