A certain heat engine operating on a Carnot cycle absorbs 150 J of heat per cycle at its hot res

A certain heat engine operating on a Carnot cycle absorbs...

A certain heat engine operating on a Carnot cycle absorbs 150 $\mathrm{J}$ of heat per cycle at its hot reservoir at $135^{\circ} \mathrm{C}$ and has a thermal efficiency of 22.0$\%$ . (a) How much work does this engine do per cycle? (b) How much heat does the engine waste each cycle? (c) What is the temperature of the cold reservoir? (d) By how much does the engine change the entropy of the world each cycle? (e) What mass of water could this engine pump per cycle from a well 35.0 $\mathrm{m}$ deep?

Key Concepts

Carnot Cycle

The Carnot cycle is an idealized thermodynamic cycle that defines the maximum possible efficiency any heat engine can achieve when operating between two thermal reservoirs. It consists of two isothermal processes (where heat transfer occurs at constant temperature) and two adiabatic processes (where no heat is exchanged). The Carnot cycle serves as a benchmark for comparing real engines, and its reversible nature implies no net change in entropy for the complete cycle of the system.

Thermal Efficiency

Thermal efficiency measures the fraction of heat absorbed by a heat engine that is converted into work. In theoretical terms, especially for a Carnot engine, it is determined solely by the temperatures of the hot and cold reservoirs. This concept is crucial for understanding how effectively an engine uses heat to perform work and for calculating the work output and waste heat when the efficiency is known.

Work and Heat Transfer

In thermodynamic systems like heat engines, energy conservation dictates that the work output plus the waste heat must equal the heat absorbed from the hot reservoir. This relationship is essential in determining how much work is produced and how much energy is rejected to the cold reservoir. Understanding this balance allows for the analysis of engine performance and energy transfer in each operational cycle.

Entropy Change

Entropy is a measure of the disorder or randomness within a system and is a key concept in the second law of thermodynamics. In a complete reversible cycle such as the Carnot cycle, the total entropy change of the system is zero, though individual components (the reservoirs) may experience changes. The concept is used to assess the irreversibility of real processes by comparing them with the ideal reversible process, and it allows us to evaluate the impact of the engine's operation on the overall entropy, or 'disorder', of the universe.

Gravitational Potential Energy

Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. In the context of applications such as pumping water, the energy required to lift water to a certain height is determined by the product of the water's mass, the gravitational acceleration, and the height. This concept connects thermodynamic work output from a heat engine to practical mechanical work, such as elevating water against gravity.

Transcript

00:02 Given absorbed heat by kornot engine that it is represented by qh which is 150 jule per cycle temperature of hot reservoir let it is represented by th which is equal to 135 degrees centigrade now we will convert into kelvin so 135 plus 273 now it is converted into kelvin which is equal to 200, 408 kelvin.

00:48 Thermal efficiency, let it is represented by e, which is given 22%.

00:55 22 .0%.

00:58 So we have to find the work done by the heat engine per cycle.

01:06 Let it is represented by w.

01:09 So we have to find the work done by heat engine per cycle.

01:13 So efficiency of kornot indian, we know that efficiency of kornat indian is equal to w upon qh.

01:23 So from this we can calculate w is equal to e into qh.

01:32 Now we will substitute the value of e and qh.

01:36 It is given 22%.

01:39 Efficiency is 22 % so 22 % of 100 into qh is 100.

01:46 So from this after calculation we will get 33 jules now work done by heat engine per cycle is equal to 33 jolt now we will look into the bit b waste of heat engine we have to find the wastage of heat engine wastage of energy by the heat engine qh qc we have to find we know that qc is equal to qh minus w for kornot engine.

02:29 So from this we will substitute the value of qc of qh and w.

02:36 Qh is given we have calculated 33 joules from the bit a minus 150 which is equal to 100 minus 117 joule.

02:55 Qc is equal to minus 117 jouc is equal to minus 117 now for the bit c, let the temperature of cold reservoir, it is represented by tc.

03:16 So here we have to find the temperature of cold reservoir.

03:20 We know that qc upon qh is equal to minus tc upon th.

03:30 Now we will substitute the value.

03:33 So from this we will get tc is equal to minus t .h.

03:39 Into qc upon qh.

03:43 Now we will substitute the value in this...