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Entropy Change from Digesting Fat. Digesting fat produces 9.3 food calories per gram of fat, and typically 80\(\%\) of this energy goes to heat when metabolized. (One food calorie is 1000 calories and therefore equals 4186 \(\mathrm{J} . )\) The body then moves all this heat to the surface by a combination of thermal conductivity and motion of the blood. The internal temperature of the body (where digestion occurs) is normally \(37^{\circ} \mathrm{C},\) and the surface is usually about \(30^{\circ} \mathrm{C} .\) By how much do the digestion and metabolism of a \(2.50-\mathrm{g}\) pat of butter change your body's entropy? Does it increase or decrease?

Short Answer

Expert verified
The body's entropy increases by 6.11 J/K.

Step by step solution

01

Convert Food Calories to Joules

First, we need to convert the energy produced from digesting the fat into joules. We have \(9.3 \) food calories per gram. Therefore, for a \(2.50\) g pat of butter, the energy in food calories is \(2.50 \times 9.3 = 23.25\) food calories. Since \(1\) food calorie equals \(4186\) J, the total energy in joules is \[23.25 \times 4186 = 97246.5\text{ J}.\]
02

Calculate the Energy Converted to Heat

Typically \(80\%\) of the metabolized energy goes to heat. Therefore, the energy in joules converted to heat is \(0.80 \times 97246.5 \text{ J} = 77797.2\text{ J}.\)
03

Determine Temperature in Kelvin

Convert the digestion and surface temperatures from Celsius to Kelvin: \[T_{internal} = 37^{\circ}\text{C} + 273.15 = 310.15\text{ K}\] and \[T_{surface} = 30^{\circ}\text{C} + 273.15 = 303.15\text{ K}.\]
04

Calculate the Change in Entropy

The change in entropy \(\Delta S\) is calculated using the formula: \[\Delta S = \frac{Q}{T_{surface}} - \frac{Q}{T_{internal}},\] where \(Q\) is the heat energy in joules.So, substituting the values we have: \[\Delta S = \frac{77797.2}{303.15} - \frac{77797.2}{310.15}.\]Performing the calculations: \[\Delta S = 256.80 - 250.69 = 6.11 \text{ J/K}.\]
05

Determine if Entropy Increases or Decreases

Since the change in entropy \(\Delta S\) is positive \(6.11 \text{ J/K},\) the entropy of the system increases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
Thermodynamics is a branch of physics that deals with heat, energy, and work. It explores how energy is transformed within a system and its surroundings. Here, the digestion of fat is an excellent example of thermodynamics at work. When the body metabolizes fat, it converts the chemical energy stored in fat molecules into heat energy. This transformation follows the laws of thermodynamics. Specifically, it involves energy conservation, which states that energy cannot be created or destroyed, only transformed. In this context, when the body converts calories from ingested fat into heat, it's essentially turning caloric energy into thermal energy. The first law of thermodynamics is evident here as it mandates that the total energy remains constant. But how does this affect the body's functions and processes? That's where the interplay of caloric energy conversion, metabolic processes, and entropy calculation come into play.
Caloric Energy Conversion
Caloric energy conversion is the process through which the body converts calories from food into usable energy. When you eat a piece of butter, your body uses the chemical reactions in digestion to release this caloric energy. This energy initially measured as food calories, is converted into joules, the unit for energy in physics. For example, each food calorie equals 4186 joules. During this conversion, the energy derived can either be used for mechanical work or dissipated as heat. In the case of fat metabolism, about 80% of energy is released as heat. This release of heat is crucial for maintaining body temperature and supporting other bodily functions like blood circulation. The efficiency of this conversion process affects how much of the consumed energy is available to perform work or be stored.
Metabolic Processes
Metabolic processes are complex biochemical reactions that convert the nutrients we consume into energy and building blocks for growth, reproduction, and maintenance. These processes are vital for sustaining life and ensuring that our bodies have the energy they need. When we consume fat, enzymes break it down into smaller components. This breakdown of fat releases energy, which the body uses or stores. The process not only involves direct conversion but also entails transporting the energy-rich molecules to where they are needed. Ultimately, the conversion of fat to energy is an ongoing metabolic process. It involves several steps, including digestion, absorption, transportation, and conversion into energy. Each step is controlled by specific enzymes and regulates how efficiently the body can use the consumed fat. As the body adjusts to this metabolic activity, it also deals with the generation and dissipation of heat, impacting entropy levels.
Entropy Calculation
Entropy is a measure of disorder or randomness in a system. In thermodynamics, it's crucial to understand how changes in energy affect a system's entropy. When digesting fat, entropy change helps quantify how the energy transformation impacts the body. The body, while converting fat calories to heat, experiences an increase in entropy. This is calculated by subtracting the heat absorbed at the body's surface temperature from the heat released at the internal temperature. Using the formula: \(\Delta S = \frac{Q}{T_{surface}} - \frac{Q}{T_{internal}}\) Where \(Q\) is the heat energy in joules, this calculation supplies insight into how energy disperses, and why an increase in energy randomness corresponds with an increase in entropy. In the exercise, as heat moves from a higher internal temperature to a cooler surface temperature, the entropy rises, indicating a natural progression toward greater disorder. This increase is fundamental to understanding not only the digestion of food but the broader principles of heat transfer and energy flow.

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Most popular questions from this chapter

A refrigerator has a coefficient of performance of \(2.10 .\) In each cycle it absorbs \(3.40 \times 10^{4} \mathrm{J}\) of heat from the cold reservoir.(a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?

You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 \(\mathrm{kg}\) of ice. During this time, how much work \(W\) is performed by the engine?

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured-that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

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