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Chapter 20: Problem 25

If a person does \(8.00 \mathrm{~h}\) of moderate physical labor "burning" 400 kcal/h, by how much does his or her internal energy change as a result?

Short Answer

Expert verified
The internal energy decreases by 13388800 J.

Step by step solution

01

Determine Total Calories Burned

To find out how many total calories the person has burned, multiply the rate of calorie burning by the number of hours: \[\text{Calories burned} = 8.00 \, \text{hours} \times 400 \, \text{kcal/hour} = 3200 \, \text{kcal}\]
02

Convert Calories to Joules

To convert the calories burned to joules, use the conversion factor where 1 kcal equals 4184 joules. Hence,\[\text{Energy in joules} = 3200 \, \text{kcal} \times 4184 \, \text{J/kcal} = 13388800 \, \text{J}\]
03

Determine Change in Internal Energy

The change in internal energy is equal to the energy burned during the physical labor. Since the person has expended this energy, their internal energy decreases by this amount:\[\Delta U = -13388800 \, \text{J}\]

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Key Concepts

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

Calories to Joules Conversion
When talking about energy expenditure in the body, calories are often the first unit that comes to mind. However, in many scientific contexts, energy is measured in joules. To make it easier to understand energy changes in physical processes, converting calories to joules is a common practice.
One dietary calorie (kcal) is equivalent to 4184 joules. This conversion is crucial for analyzing tasks where energy units need to be consistent, such as in physics or engineering.
To convert calories to joules, simply multiply the number of calories by 4184. For example, if a person burns 3200 kcal during a workout, this would equal 3200 kcal 脳 4184 J/kcal = 13388800 joules.
  • This helps us comprehend how much energy is being used at a fundamental level.
  • Understanding this conversion is key in bridging nutritional science with physics-related calculations.
Energy Expenditure
Energy expenditure refers to the amount of energy an individual uses to perform physical activities. This includes everyday activities like walking, running, or exercise.
The amount of energy expended varies significantly depending on the type of activity, its intensity, and the individual's body weight. In our specific case, the energy expenditure was calculated based on the person's physical labor over a span of hours.
Everyday life, as much as physical labor, requires careful energy management. For instance, when someone engages in moderate labor for 8 hours burning 400 kcal per hour, they expend a lot of energy, impacting their internal energy reserve.
  • Tracking energy expenditure can help optimize dietary and fitness routines.
  • Balance between caloric intake and energy expenditure is essential for maintaining or changing body weight.
Physical Labor Energy Usage
Engaging in physical labor requires significant energy, as it involves muscle activity that rapidly consumes calories. The human body's energy expenditure increases with the intensity of labor.
Physical labor can vary from light activities, like sweeping or gardening, to more intense ones like heavy lifting or construction work. For moderate labor, the average rate can be around 400 kcal per hour.
Such an extensive activity over multiple hours, such as working 8 hours straight, results in considerable energy usage and consequently a decline in the body's internal energy levels.
  • Regular physical labor demands substantial caloric intake to replenish energy levels.
  • Energy usage in labor is an important aspect of occupational health and fitness planning.
Thermodynamics in Human Body
Thermodynamics, often associated with engines and machines, also applies to human bodies. Our bodies are complex systems that transform food into energy using thermodynamic principles.
When a person performs physical labor, the law of conservation of energy is at play. The food consumed provides calories, which translate into energy produced and expended during activities.
The internal energy of a person decreases as calories are used in physical tasks鈥攁 reflection of thermodynamic changes within the body. Emphasizing energy input and output balancing is key to understanding the body's efficiency.
  • Metabolic processes illustrate how energy is managed and mechanistically controlled.
  • Efficiency of energy use affects overall health and performance in physical activities.

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Most popular questions from this chapter

A 70 -g metal block moving at \(200 \mathrm{~cm} / \mathrm{s}\) slides across a tabletop a distance of \(83 \mathrm{~cm}\) before it comes to rest. Assuming 75 percent of the thermal energy developed by friction goes into the block, how much does the temperature of the block rise? For the metal, \(c=\) \(0.106 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\)

A sealed chamber containing \(32.5\) g of molecular oxygen and \(20.2\) g of molecular nitrogen at \(48.0^{\circ} \mathrm{C}\) is cooled down to \(20.2^{\circ} \mathrm{C}\). Given that for \(\mathrm{N}_{2}, c_{v}=0.743 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), and for \(\mathrm{O}_{2}, c_{v}=0.659\) \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the resulting change in the internal energy of the gas.

(a) Compute \(c_{v}\) for the monatomic gas argon, given \(c_{p}=0.125\) cal/g \(\cdot{ }^{\circ} \mathrm{C}\) and \(\mathrm{y}=1.67 .\) (b) Compute \(c_{p}\) for the diatomic gas nitric oxide \((\mathrm{NO})\), given \(c_{v}=0.166 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(\mathrm{Y}=1.40\).

The temperature of \(5.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}\) gas is raised from \(10.0{ }^{\circ} \mathrm{C}\) to \(130.0{ }^{\circ} \mathrm{C}\). If this is done at constant volume, find the increase in internal energy \(\Delta U\). Alternatively, if the same temperature change now occurs at constant pressure determine both \(\Delta V\) and the external work \(\Delta W\) done by the gas. For \(\mathrm{N}_{2}\) gas, \(c_{v}=0.177\) cal \(/ \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(c_{p}=0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If the gas is heated at constant volume, then no work is done during the process. In that case \(\Delta W=0\), and the First Law tells us that \((\Delta Q)_{v}=\Delta U\). Because \((\Delta Q)_{v}=c_{v} m \Delta T\), \(\Delta U=(\Delta Q)_{v}=\left(0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)=106 \mathrm{kcal}=443 \mathrm{~kJ}\) The temperature change is a manifestation of the internal energy change. When the gas is heated \(120^{\circ} \mathrm{C}\) at constant pressure, the same change in internal energy occurs. In addition, however, work is done. The First Law then becomes $$(\Delta Q)_{\mathrm{p}}=\Delta U+\Delta W=443 \mathrm{~kJ}+\Delta W$$ But \((\Delta Q)_{\mathrm{p}}=c_{p} m \Delta T=\left(0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)\) $$=149 \mathrm{kcal}=623 \mathrm{~kJ}$$ Hence \(\Delta W=(\Delta Q)_{\mathrm{p}}-\Delta U=623 \mathrm{~kJ}-443 \mathrm{~kJ}=180 \mathrm{~kJ}\)

A gas at a pressure of \(2.10 \times 10^{5}\) Pa occupies \(4.98 \times 10^{-3} \mathrm{~m}^{3}\) in a chamber that can change its volume. The gas is at an initial temperature of \(290 \mathrm{~K}\) when it is heated, so that it expands isobarically, thereupon doing \(200 \mathrm{~J}\) of work. Determine the new volume and the final temperature of the gas. [Hint: Use Eq. (20.1) and the Ideal Gas Law applied before and after the expansion.]
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