/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 You have no doubt noticed that y... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C}\). and it produces approximately \(290 \mathrm{~W}\) of heat power per square meter of body area. A \(68 \mathrm{~kg}(150 \mathrm{lb}), 1.78 \mathrm{~m}(5 \mathrm{ft}, 10\) in.) person has approximately \(1.8 \mathrm{~m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat of the body is about \(3500 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}).\)

Short Answer

Expert verified
The person has to shiver for about 7.6 minutes.

Step by step solution

01

Calculate the Total Heat Needed

To calculate the total heat required to raise the body temperature by 1.0°C, use the formula \( Q = mc\Delta T \). Where \( m \) is the mass (68 kg), \( c \) is the specific heat capacity (3500 J/kg·K), and \( \Delta T \) is the temperature change (1.0°C or 1.0 K, since the change in Celsius and Kelvin is the same).\[ Q = 68 \times 3500 \times 1.0 = 238,000 \text{ J} \]
02

Calculate the Heat Power Produced

Now, calculate the total heat power produced by the person shivering. The heat power generated per square meter is 290 W. If the surface area is 1.8 m², the total power is:\[ P = 290 \times 1.8 = 522 \text{ W} \]
03

Calculate Time Required to Generate Required Heat

To find out how long it takes to generate the required amount of heat, use the formula \( Q = Pt \), where \( P \) is power and \( t \) is time. Rearrange the formula to solve for time:\[ t = \frac{Q}{P} \]Substitute the known values:\[ t = \frac{238,000}{522} \approx 456 \text{ seconds} \]
04

Convert Time to Minutes

Convert the time from seconds to minutes by dividing by 60:\[ t \approx \frac{456}{60} \approx 7.6 \text{ minutes} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Specific Heat Capacity
Understanding specific heat capacity is crucial when studying thermodynamics. It's the measure of how much heat energy is needed to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). For example, the specific heat capacity of the human body is about 3500 J/(kg·K).
This means that 3500 joules of energy is needed to raise the temperature of one kilogram of body mass by 1°C.
Different substances have different specific heat capacities, affecting how quickly they heat up or cool down.
  • Water, for example, has a high specific heat capacity, meaning it can absorb a lot of heat without a significant change in temperature.
  • This property plays a key role in energy transfer and temperature regulation in many biological and physical processes.
Heat Transfer
Heat transfer is the process of thermal energy moving from a hotter object to a cooler one. In this context, when you're shivering, your body is transferring internal energy to keep warm.
There are three primary methods of heat transfer:
  • Conduction is direct heat transfer between substances in contact, such as your feet on cold tiles.
  • Convection involves the movement of heat through fluids, like the warmth of a hot drink spreading in your stomach.
  • Radiation transfers heat via electromagnetic waves, such as sunlight warming your skin.
These processes explain how energy is distributed, affecting how quickly or slowly an item heats up. Your body uses these methods to maintain its internal temperature efficiently.
Energy Conservation
Energy conservation in thermodynamics deals with how energy is neither created nor destroyed, only transformed or transferred.
In the exercise, you generate heat through shivering, converting chemical energy from your body into thermal energy.
  • This ensures energy remains in balance, following the first law of thermodynamics.
  • The heat produced by shivering is used to raise your body temperature.
  • Through energy conversion, your body's mechanisms work to maintain homeostasis, even in cooling conditions.
By understanding energy conservation, it becomes clear how dynamically energy flows through different systems, helping organisms adapt to various temperatures.
Body Temperature Regulation
Body temperature regulation is a vital process to ensure the body works correctly. The average human body temperature is around 37°C, and it is crucial to maintain it within a narrow range.
When you step out of a shower and shiver, your body responds by generating heat to counteract heat loss and maintain this stable internal environment.
  • Shivering is an involuntary response that generates heat through muscle activity.
  • Other body responses include altering blood flow, adjusting metabolic heat production, and employing clothing or behavior to conserve heat.
  • These mechanisms ensure the body's core temperature remains stable, safeguarding vital functions.
Understanding how the body regulates temperature provides insight into how it maintains an energy balance through thermoregulation processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Maintaining body temperature. While running, a \(70 \mathrm{~kg}\) student generates thermal energy at a rate of \(1200 \mathrm{~W}\). To maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Protein structures in the body are damaged irreversibly if the body temperature rises to \(44^{\circ} \mathrm{C}\) or above. The specific heat of a typical human body is \(3480 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}),\) slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

(a) At what temperature do the Fahrenheit and Celsius scales give the same reading? (b) Is there any temperature at which the Kelvin and Celsius scales coincide?

for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body's temperature can go as high as \(40^{\circ} \mathrm{C}\). What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about \(7 \mathrm{C}^{\circ}\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at \(4.0^{\circ} \mathrm{C}\) lasts safely for about 3 weeks, whereas blood stored at \(-160^{\circ} \mathrm{C}\) lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body's temperature is above \(105^{\circ} \mathrm{F}\) for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course. be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

Each pound of fat contains 3500 food calories. When the body metabolizes food, \(80 \%\) of this energy goes to heat. Suppose you decide to run without stopping, an activity that produces \(1290 \mathrm{~W}\) of metabolic power for a typical person. (a) For how many hours must you run to burn up 1 lb of fat? Is this a realistic exercise plan? (b) If you followed your planned exercise program, how much heat would your body produce when you burn up a pound of fat? (c) If you needed to get rid of all of this excess heat by evaporating water (i.e., sweating), how many liters would you need to evaporate? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.