A Carnot engine extracts heat from a block of mass \(m\) and specific heat \(c\) initially at temperature \(T_{\mathrm{h} 0}\) but without a heat source to maintain that temperature. The engine rejects heat to a reservoir at constant temperature \(T_{\mathrm{c}} .\) The engine is operated so its mechanical power output is proportional to the temperature difference \(T_{\mathrm{h}}-T_{\mathrm{c}}\) : $$P=P_{0} \frac{T_{\mathrm{h}}-T_{\mathrm{c}}}{T_{\mathrm{h} 0}-T_{\mathrm{c}}}$$ where \(T_{\mathrm{h}}\) is the instantaneous temperature of the hot block and \(P_{0}\) is the initial power. (a) Find an expression for \(T_{\mathrm{h}}\) as a function of time, and (b) determine how long it takes for the engine's power output to reach zero.

The heart of a heat engine, like the Carnot engine, lies in the heat engine cycle. This process involves converting heat energy into mechanical work by transferring heat between two reservoirs at different temperatures. Here's a simplified breakdown of the cycle:

• First, the engine absorbs heat from the hot reservoir, which causes the working substance, typically a gas, to expand and do work on the surroundings.
• The engine then performs adiabatic (no heat exchange) expansion, further converting thermal energy to work.
• Next, the engine releases heat to the cold reservoir while the working substance contracts, causing a decrease in internal energy.
• Finally, adiabatic compression restores the initial state of the working substance, completing the cycle.

The Carnot cycle is an idealized version of the heat engine cycle and operates with maximum possible efficiency. It was imagined by Sadi Carnot, who was trying to understand the limits of efficiency for heat engines. A key result from his analysis was that the efficiency depends only on the temperatures of the hot and cold reservoirs. For students wrestling with these concepts, remembering that a cycle involves stages of heat exchange and mechanical work, often visualized on a pressure-volume diagram, is crucial.

In the study of thermodynamics, an essential concept is the specific heat capacity. This property is a measure of how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It's represented by the symbol 'c' and typically expressed in units of J/(kg⋅K).

What does this mean in the context of our exercise? When dealing with a Carnot engine, the specific heat of the substance from which the engine extracts heat will determine how quickly its temperature can change. This is also crucial in calculating the rate at which the engine loses heat to the environment, as outlined in Step 2 of the solution. The higher the specific heat capacity, the more heat is required to change the temperature, and vice versa.

In our exercise, we encounter a differential equation that emerges from applying the principles of thermodynamics to the Carnot engine. This equation represents a fundamental relationship between several variables: time, temperature, and heat transfer rate—and provides a way to describe how these quantities evolve over time.

Solving differential equations, like the one in Step 4, is a critical skill in physics and engineering. The approach involves separating variables to isolate the differential terms and then integrating both sides to find the desired function—here, the temperature as a function of time. To simplify this concept for students, one can think of it akin to unraveling a puzzle: each move (integration step) reveals part of the picture (the solution function), until you finally see the full image (temperature change over time).

A Carnot engine's power output is central to our discussion and the exercise at hand. Power in a thermodynamic context is the rate at which work is performed. For a Carnot engine, the power output is directly linked to the efficiency of the cycle and the rate of heat transfer from the hot reservoir. The equation provided in the exercise illustrates that as the hot reservoir cools, the power output decreases because it's proportional to the temperature difference between the hot reservoir and the cold one.

To better understand the power output for students, correlating with Step 5, consider what power output practically means: it's like the energy per second that the engine could use to perform work—such as turning the wheels of a train. Thus, as the temperature of the hot block approaches the cold reservoir's temperature, the energy available for work per second decreases until, eventually, the engine cannot do any more work, and the power output reaches zero.