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Chapter 17: Problem 36

Heat Loss During Breathing. In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C}\) , what amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat of air is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that 1.0 \(\mathrm{L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{kg}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

Short Answer

Expert verified
37.719 J per breath; 45,262.8 J per hour.

Step by step solution

01

Convert Temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin, since the formula for heat transfer uses temperatures in Kelvin. \(-20^{\circ}\, C\) becomes \(-20 + 273.15 = 253.15\, K\) and \(37^{\circ}\, C\) becomes \(37 + 273.15 = 310.15\, K\).
02

Calculate Temperature Change

Now, find the change in temperature \(\Delta T\). It is the difference between the final and initial temperatures in Kelvin: \(\Delta T = 310.15\, K - 253.15\, K = 57\, K\).
03

Determine Mass of Air per Breath

With the given density of air, calculate the mass of 0.50 L of air. This is done using: \(\text{mass} = \text{volume} \times \text{density} = 0.50\, \text{L} \times 1.3 \times 10^{-3}\, \text{kg/L} = 6.5 \times 10^{-4}\, \text{kg}\).
04

Calculate Heat Required per Breath

Use the formula for heat transfer: \(Q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the temperature change. \(Q = 6.5 \times 10^{-4}\, \text{kg} \times 1020\, \text{J/kg}\, K \times 57\, K = 37.719\, J \).
05

Calculate Heat Lost per Minute

Multiply the heat per breath by the respiration rate to find the heat lost per minute. \(Q_{per\, minute} = 37.719\, J \times 20 = 754.38\, J/min\).
06

Calculate Heat Lost per Hour

Multiply the heat lost per minute by 60 to convert to heat lost per hour. \(Q_{per\, hour} = 754.38\, J/min \times 60\, min = 45262.8\, J\).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is an intrinsic property of substances that measures the amount of heat per unit mass required to raise the temperature by one degree Celsius (or one Kelvin). In our scenario, air has a specific heat capacity of 1020 J/kg路K, meaning for every kilogram of air, 1020 Joules are needed to raise its temperature by 1 Kelvin. This property is crucial for calculating heat transfer as it directly influences how much energy is needed to change the temperature of the air we inhale from a cold temperature to body temperature.
  • Helps understand energy changes in thermal scenarios.
  • Essential for calculating energy requirements in processes such as breathing in cold environments.
Heat Transfer
Heat transfer is the process by which thermal energy moves from a warmer object or region to a cooler one. In the context of breathing, this involves the body transferring heat to the cold air we inhale during winter to warm it up to body temperature. The calculation of heat transfer involves using the formula:\[Q = mc\Delta T\]where:
  • \( Q \) is the heat transferred (in Joules),
  • \( m \) is the mass of the air breathed in (in kg),
  • \( c \) is the specific heat capacity (J/kg路K),
  • \( \Delta T \) is the change in temperature (in K).
By understanding how heat is transferred, one can appreciate the body's energy expenditure in maintaining an optimal temperature for processes like respiration.
Respiration Rate
The respiration rate, or breathing rate, refers to the number of breaths a person takes per minute. It plays a significant role in calculating the total heat lost by the body due to breathing. This rate affects the total volume of air inhaled and exhaled, thereby influencing the overall heat that needs to be transferred. For this exercise, a respiration rate of 20 breaths per minute is considered. This means that over the course of an hour, the number of breaths will increase the total energy expenditure:
  • Multiply the energy required per breath with the respiration rate to find the total heat expenditure over a given time period, such as one minute or one hour.
  • This contributes to understanding how physical activity and temperature impact energy loss during respiration in cold weather.
Temperature Conversion
In thermodynamics, temperature conversion is often needed to facilitate calculations that require temperature differences. While humans generally use the Celsius scale, scientific calculations with heat transfer prefer Kelvin. The conversion between Celsius and Kelvin is straightforward: add 273.15 to the Celsius temperature to convert it to Kelvin.
  • This ensures that temperature differences, which are the same for Celsius and Kelvin, are accurately represented in calculations.
  • Using Kelvin helps avoid negative values when working with temperature changes, which is particularly useful when calculating work and energy transfer.
For example, in this case, converting the temperatures to Kelvin before finding the temperature difference ensures that all subsequent calculations for heat transfer use the correct values.

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Most popular questions from this chapter

Two beakers of water, \(A\) and \(B\) , initially are at the same temperature. The temperature of the water in beaker \(A\) is increased \(10 F^{\circ},\) and the temperature of the water in beaker \(B\) is increased 10 \(\mathrm{K}\) . After these temperature changes, which beaker of water has the higher temperature? Explain.

"The Ship of the Desert" Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 \(\mathrm{kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \(\left(12 \text { hours) instead of letting it rise to } 40.0^{\circ} \mathrm{C} \text { . (Note: The }\right.\) specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) )

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ}\) ) above the temperature of the outer surface. The total area of the box is 2.18 \(\mathrm{m}^{2}\) , and the wall thickness is 3.90 \(\mathrm{cm}\) . Find the thermal conductivity of the material in SI units.

Spacecraft Reentry. A spacecraft made of aluminum circles the earth at a speed of 7700 \(\mathrm{m} / \mathrm{s}\) . (a) Find the ratio of its kinetic energy to the energy required to raise its temperature from \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\) . (The melting point of aluminum is \(660^{\circ} \mathrm{C}\) . Assume a constant specific heat of \(910 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) (b) Discuss the bearing of your answer on the problem of the reentry of a manned space vehicle into the earth's atmosphere.

Before going in for his annual physical, a \(70.0-\mathrm{kg}\) man whose body temperature is \(37.0^{\circ} \mathrm{C}\) consumes an entire \(0.355-\mathrm{L}\) can of a soft drink (mostly water) at \(12.0^{\circ} \mathrm{C}\) . (a) What will his body temperature be after equilibrium is attained? Ignore any heating bythe man's metabolism. The specific heat of the man's body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . (b) Is the change in his body temperature great enough to be measured by a medical themometer?
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