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Chapter 16: Problem 28

A typical doughnut contains approximately 200 food calories (kilocalories), of which about \(80 \%\) is metabolized into heat by your body. Assume that the internal temperature of your body (where digestion occurs) is normally \(37^{\circ} \mathrm{C}\) and that this does not significantly change after you digest the doughnut. By how much does your body's entropy change after you eat a doughnut?

Short Answer

Expert verified
The body's entropy increases by approximately 2,156.8 J/K after eating the doughnut.

Step by step solution

01

Understand the problem

We are asked to find the change in entropy of the body after metabolizing a doughnut. We know the energy content of the doughnut, the efficiency of conversion to heat, and the body's internal temperature.
02

Calculate the usable energy

A typical doughnut contains 200 kcal. Since 80% of this energy is metabolized into heat, we calculate the usable heat energy as follows: \[\text{Usable energy} = 0.8 \times 200 \text{ kcal} = 160 \text{ kcal} = 160,000 \text{ cal} = 160,000 \times 4.184 \text{ J} = 668,800 \text{ J} \]
03

Recall the formula for entropy change

The entropy change (\( \Delta S \)) when heat (\( Q \)) is transferred at a constant temperature (\( T \)) is given by:\[\Delta S = \frac{Q}{T}\] where \(T\) must be in Kelvin.
04

Convert the temperature to Kelvin

The body's internal temperature is given in Celsius as \(37^{\circ} \mathrm{C}\). Convert this temperature to Kelvin:\[T = 37 + 273.15 = 310.15 \text{ K}\]
05

Calculate the change in entropy

Using the formula for entropy change and the calculated values:\[\Delta S = \frac{668,800 \text{ J}}{310.15 \text{ K}} = 2,156.8 \text{ J/K} \]
06

Conclusion

The change in the body's entropy after metabolizing the doughnut is approximately 2,156.8 J/K.

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Key Concepts

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

Energy Metabolism
Energy metabolism refers to the processes our bodies use to convert the food we eat into usable energy. This energy is essential for sustaining life, enabling us to move, think, and even breathe.
  • When we eat, our body breaks down the food into its basic components: carbohydrates, fats, and proteins.
  • These components are then transformed into energy through metabolic pathways.
  • Most importantly, a significant portion of this energy transformation results in heat, which helps maintain body temperature.
In the case of our exercise, a doughnut provides 200 kilocalories, with 80% of this energy being converted to heat. Thus, energy metabolism not only supports our physical activities but also plays a crucial role in maintaining our body's thermal homeostasis.
Thermodynamics
Thermodynamics is the branch of physics dealing with heat, work, and the forms of energy transfer. It provides a theoretical framework to understand processes where heat is converted into mechanical work or other energy forms.
  • The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another.
  • The second law introduces the concept of entropy, a measure of disorder or randomness within a system.
  • Understanding these laws helps in analyzing processes like energy metabolism within the human body.
In our problem, thermodynamics principles aid in calculating entropy changes as the doughnut's energy undergoes conversion into body heat.
Kelvin Temperature Scale
The Kelvin temperature scale is a vital concept in scientific fields, as it provides an absolute measure of temperature. Unlike Celsius or Fahrenheit, Kelvin starts at absolute zero, the point where all molecular motion ceases.
  • 1 Kelvin degree is equivalent in magnitude to 1 degree Celsius, but they differ in their starting points.
  • To convert Celsius to Kelvin, simply add 273.15.
In the given exercise, the internal body temperature of 37掳C is converted to 310.15 K to use in entropy calculations. Utilizing the Kelvin scale ensures that we are working within the context of thermodynamics laws, which are best expressed using absolute temperature measures.
Heat Transfer
Heat transfer is the process by which thermal energy is exchanged between physical systems due to temperature differences. There are three primary modes of heat transfer: conduction, convection, and radiation.
  • Conduction involves heat transfer through direct contact between materials.
  • Convection is the transfer of heat through fluids (gases or liquids) by the motion of those fluids.
  • Radiation involves the transfer of heat via electromagnetic waves, without needing a medium.
In biological systems, heat transfer is essential to maintaining body temperature and facilitating metabolic reactions. As in our task, the body's metabolism of a doughnut into heat is a form of internal energy transfer, highlighting the body's efficiency in converting stored chemical energy into thermal energy.

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Most popular questions from this chapter

A \(4.50 \mathrm{~kg}\) block of ice at \(0.00^{\circ} \mathrm{C}\) falls into the ocean and melts. The average temperature of the ocean is \(3.50^{\circ} \mathrm{C},\) including all the deep water. By how much does the melting of this ice change the entropy of the world? Does it make it larger or smaller? (Hint: Do you think that the ocean will change temperature appreciably as the ice melts?)

A Carnot engine has an efficiency of \(59 \%\) and performs \(2.5 \times 10^{4} \mathrm{~J}\) of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle? (b) Suppose the engine exhausts heat at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) What is the temperature of its heat source?

A window air-conditioner unit absorbs \(9.80 \times 10^{4} \mathrm{~J}\) of heat per minute from the room being cooled and in the same period deposits \(1.44 \times 10^{5} \mathrm{~J}\) of heat into the outside air. What is the power consumption of the unit in watts?

A Carnot engine whose high-temperature reservoir is at \(620 \mathrm{~K}\) takes in \(550 \mathrm{~J}\) of heat at this temperature in each cycle and gives up \(335 \mathrm{~J}\) to the low-temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? (b) What is the temperature of the low-temperature reservoir? (c) What is the thermal efficiency of the cycle?

In one cycle, a freezer uses \(785 \mathrm{~J}\) of electrical energy in order to remove \(1750 \mathrm{~J}\) of heat from its freezer compartment at \(10^{\circ} \mathrm{F}\). (a) What is the coefficient of performance of this freezer? (b) How much heat does it expel into the room during this cycle?
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