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The average heat evolved by the oxidation of foodstuffs in an average adult per hour per kilogram of body weight is \(7.20 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{hr}^{-1}\). Assume the weight of an average adult is \(62.0 \mathrm{kg} .\) Suppose the total heat evolved by this oxidation is transferred into the surroundings over a period lasting one week. Calculate the entropy change of the surroundings associated with this heat transfer. Assume the surroundings are at \(T=293 \mathrm{K}\)

Short Answer

Expert verified
To calculate the entropy change of the surroundings associated with the heat transfer over one week, first find the heat transferred per hour: \( Q_{hour} = 62.0 \mathrm{kg} \times 7.20 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{hr}^{-1} \). Next, calculate the heat transferred over one week: \( Q_{week} = Q_{hour} \times 168 \mathrm{hrs} \). Finally, calculate the entropy change using \( \Delta S = \frac{Q_{week}}{T} \), with T = 293 K.

Step by step solution

01

Calculate the Heat Transferred Per Hour

To calculate the heat transferred per hour, we will use the heat evolved per hour and the weight of the adult, given as: Weight of adult: \( 62.0 \mathrm{kg} \) Heat evolved per hour per kilogram: \( 7.20 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{hr}^{-1} \) Heat transferred per hour, \( Q_{hour} = \) Weight × Heat evolved per hour per kilogram \( Q_{hour} = 62.0 \mathrm{kg} \times 7.20 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{hr}^{-1} \)
02

Calculate the Heat Transferred Over One Week

Now, we will calculate the heat transferred over a week by multiplying the heat transferred per hour by the total number of hours in one week. Hours in a week: \( 24 \mathrm{hrs} \times 7 = 168 \mathrm{hrs} \) Total heat transferred, \( Q_{week} = Q_{hour} \times 168 \mathrm{hrs} \)
03

Calculate the Entropy Change of the Surroundings

To calculate the entropy change of the surroundings, we need to use the total heat transferred, \( Q_{week} \), and the temperature of the surroundings, T = 293 K. Entropy change, \( \Delta S = \frac{Q_{week}}{T} \) Considering the previous steps, we can now calculate the entropy change for the surroundings.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics. It refers to the movement of thermal energy from one body or substance to another. This transfer occurs due to a temperature difference and can happen in multiple ways, including conduction, convection, and radiation. In the case of the oxidation of foodstuffs, the body converts chemical energy into heat energy. This heat then gets transferred to the surroundings.
This process is significant because it dictates how energy moves through systems and contributes to changes in temperature and energy efficiency. In our exercise, understanding heat transfer allows us to estimate how much energy an adult releases as heat per hour due to metabolic processes.
  • Conduction: Direct heat transfer through a material.
  • Convection: Heat transfer through fluid motion.
  • Radiation: Transfer via electromagnetic waves.
These principles explain how the energy from the oxidation of foodstuffs can be transferred from the body to the surrounding environment.
Oxidation of Foodstuffs
Oxidation of foodstuffs is a biological process where the body converts food into energy. This process involves breaking down carbohydrates, fats, and proteins to release energy, often in the form of heat.
During oxidation, electrons are transferred from one molecule to another, leading to the release of energy. This energy is essential for maintaining bodily functions such as movement, growth, and temperature regulation. In the context of our problem, this process is quantified by determining the amount of heat energy released per unit of body weight per hour.
The oxidation of foodstuffs plays a crucial role in determining the metabolic rate of an individual. It explains how the food we consume is ultimately "burned" to support our daily activities, releasing heat to maintain body temperature and perform work.
Thermodynamics
Thermodynamics is the science of heat, work, and energy. It provides a framework for understanding how energy transitions between different forms and bodies. The study of thermodynamics ensures we can predict how energy moves through systems, making it crucial for many scientific and engineering fields.
In the exercise, the focus is on the first law of thermodynamics, which states that energy cannot be created or destroyed—only transformed. This principle guides the understanding of how metabolic processes convert energy from food into heat, which then affects the surrounding environment.
Basic concepts include:
  • System: The part of the universe under study.
  • Surroundings: Everything outside the system.
  • Energy: The capacity to do work.
Thermodynamics helps in predicting the amount of heat that is transferred to the surroundings during the oxidation of foodstuffs when one knows the metabolic rate and body mass.
Entropy Calculation
Entropy is a measure of disorder in a system and is a central concept in the second law of thermodynamics. Calculating entropy change can provide insight into the energy dispersal in a system.
In the problem, we calculate the entropy change to understand how the heat released by an average adult during food oxidation gets distributed into the surroundings over time. The formula for entropy change, \[\Delta S = \frac{Q}{T}\] deals with understanding how heat transfer is affected by temperature. In this formula, \(Q\) is the heat transferred, and \(T\) is the absolute temperature at which the process occurs.
This calculation is fundamental for assessing the efficiency and spontaneity of processes. In thermodynamics, a positive entropy change often indicates increased energy dispersal in the surroundings, suggesting that the process is naturally favored.

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Most popular questions from this chapter

The standard entropy of \(\mathrm{Pb}(s)\) at \(298.15 \mathrm{K}\) is \(64.80 \mathrm{J}\) \(\mathrm{K}^{-1} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(s)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, s)}{\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=22.13+0.01172 \frac{T}{\mathrm{K}}+1.00 \times 10^{-5} \frac{T^{2}}{\mathrm{K}^{2}} \\] The melting point is \(327.4^{\circ} \mathrm{C}\) and the heat of fusion under these conditions is \(4770 . \mathrm{J} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(l)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, l)}{\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}}=32.51-0.00301 \frac{T}{\mathrm{K}} \\] a. Calculate the standard entropy of \(\mathrm{Pb}(l)\) at \(725^{\circ} \mathrm{C}\). b. Calculate \(\Delta H\) for the transformation \(\mathrm{Pb}\left(s, 25.0^{\circ} \mathrm{C}\right) \rightarrow\) \(\mathrm{Pb}\left(L, 725^{\circ} \mathrm{C}\right)\)

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1.10 moles of \(\mathrm{N}_{2}\) at \(20.5^{\circ} \mathrm{C}\) and 6.20 bar undergoes a transformation to the state described by \(215^{\circ} \mathrm{C}\) and 1.75 bar. Calculate \(\Delta S\) if \\[ \begin{aligned} \frac{C_{P, m}}{\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=& 30.81-11.87 \times 10^{-3} \frac{T}{\mathrm{K}}+2.3968 \times 10^{-5} \frac{T^{2}}{\mathrm{K}^{2}} \\ &-1.0176 \times 10^{-8} \frac{T^{3}}{\mathrm{K}^{3}} \end{aligned} \\]

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