Problem 44 - University Physics with Modern Physics

Question

You decide to use your body as a Carnot heat engine. The operating gas is in a tube with one end in your mouth (where the temperature is 37.0\(^\circ\)C) and the other end at the surface of your skin, at 30.0\(^\circ\)C. (a) What is the maximum efficiency of such a heat engine? Would it be a very useful engine? (b) Suppose you want to use this human engine to lift a 2.50-kg box from the floor to a tabletop 1.20 m above the floor. How much must you increase the gravitational potential energy, and how much heat input is needed to accomplish this? (c) If your favorite candy bar has 350 food calories (1 food calorie = 4186 J) and 80% of the food energy goes into heat, how many of these candy bars must you eat to lift the box in this way?

Step-by-Step Solution

Answer

The engine's efficiency is 2.25%. It requires 0.0011 candy bars to lift the box.

Step 1: Calculate Maximum Efficiency

The formula for the maximum efficiency \( \eta_{max} \) of a Carnot engine is: \[ \eta_{max} = 1 - \frac{T_C}{T_H} \] where \( T_H \) is the temperature of the hot reservoir and \( T_C \) is the temperature of the cold reservoir. First, convert the temperatures from Celsius to Kelvin: \( T_H = 37.0 + 273.15 = 310.15 \ K \) and \( T_C = 30.0 + 273.15 = 303.15 \ K \). Now, substitute these values: \[ \eta_{max} = 1 - \frac{303.15}{310.15} \approx 0.0225 \text{ or } 2.25\% \]. This low efficiency indicates the engine would not be very useful.

Step 2: Calculate Gravitational Potential Energy Increase

The increase in gravitational potential energy \( \Delta U \) is given by \( \Delta U = mgh \), where \( m = 2.50 \ kg \), \( g = 9.81 \ m/s^2 \), and \( h = 1.20 \ m \). Calculating gives \[ \Delta U = 2.50 \times 9.81 \times 1.20 = 29.43 \text{ J} \].

Step 3: Calculate Required Heat Input

Since the efficiency \( \eta = \frac{W}{Q_H} \), we have \( Q_H = \frac{W}{\eta} \) where \( W = 29.43 \text{ J} \) is the work done. With an efficiency of 2.25\%, \[ Q_H = \frac{29.43}{0.0225} \approx 1308.00 \text{ J} \]. This is the amount of heat input required to perform the work.

Step 4: Calculate Candy Bar Requirement

Each candy bar provides 350 food calories, which is \( 350 \times 4186 = 1465100 \) J. Since only 80% of this energy contributes to heat, usable energy per bar is \( 0.8 \times 1465100 \approx 1172080 \) J. To find the number of bars needed: \[ n = \frac{1308.00}{1172080} \approx 0.0011 \]. Therefore, you need essentially no bars to physically lift the box using the engine.

Key Concepts

Thermodynamic Efficiency

Thermodynamic efficiency is a critical measure for any heat engine and is defined as the ratio of work output to the heat input. For a Carnot engine, which is an idealized heat engine with maximum possible efficiency, the formula is given by: \[ \eta_{max} = 1 - \frac{T_C}{T_H} \] where \( T_H \) is the temperature of the heat source, and \( T_C \) is the temperature of the heat sink. Before using this formula, you must convert temperatures from Celsius to Kelvin, as in the Kelvin scale, absolute zero is the starting point. The Kelvin conversion is simple: add 273.15 to the Celsius value.
In the context of using your body as a heat engine, with temperatures of 37.0°C in your mouth and 30.0°C on your skin, converting these gives \( T_H = 310.15 \ K \) and \( T_C = 303.15 \ K \). Using these values, the maximum efficiency comes out to be a very low 2.25%. Such low efficiency indicates that not much of the input heat is being converted into useful work, making it ineffective for practical purposes.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses because of its position relative to a lower position. It is given by the formula: \[ \Delta U = mgh \] where \( m \) is mass in kilograms, \( g \) is the gravitational acceleration \((9.81\, m/s^2)\), and \( h \) is the height in meters. To understand it better, imagine holding a box. When you lift the box, you increase its potential to do work against gravity, thereby increasing its gravitational potential energy.
In the problem of lifting a 2.50-kg box to a height of 1.20 m, the calculation of potential energy becomes \( \Delta U = 2.50 \times 9.81 \times 1.20 = 29.43 \text{ J} \). This represents the energy required to lift the box, which is derived from the heat engine’s conversion of heat into work.

Energy Conversion

Energy conversion is the process of changing one type of energy into another. In a heat engine, the energy conversion occurs as the engine converts thermal energy (heat) into mechanical energy (work). For any practical heat engine, the conversion efficiency is always less than 100% due to various losses and inefficiencies. The efficiency \( \eta \) of an actual engine is the ratio of work done \( W \) to the heat input \( Q_H \), expressed as: \[ \eta = \frac{W}{Q_H} \] Given the work required to lift the box \( W = 29.43 \text{ J} \) and the efficiency of 2.25%, the necessary heat input to the engine can be calculated by rearranging the formula: \[ Q_H = \frac{W}{\eta} \approx 1308.00 \text{ J} \] This highlights the significant increase required in heat input to achieve a small amount of mechanical work, reflecting the inefficiency.

Heat Input Calculation

Heat input calculation involves determining the amount of energy needed to perform a given task using a thermal engine. With low efficiency figures, like in a human-engine model, the required heat input becomes noticeably greater than the work output.
Suppose a candy bar provides 350 food calories, equivalent to \( 350 \times 4186 = 1465100 \) J of energy. However, not all of this energy can be utilized for work; specifically, only 80% of the energy, or \( 1172080 \) J, becomes available as heat. To lift the box, requiring \( 1308.00 \text{ J} \) of heat, you'd calculate the number of candy bars needed: \[ n = \frac{1308.00}{1172080} \approx 0.0011 \] This ratio implies that one does not need to consume an entire candy bar to do the work, demonstrating the challenge of human-based heat engines in practical applications. It provides a clear view of how much more input energy is required relative to the mechanical work achieved.