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While running, a \(70 \mathrm{~kg}\) student generates thermal energy at a rate of \(1200 \mathrm{~W}\). For the runner 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 energy could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. 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.)

Short Answer

Expert verified
The student could run for approximately \(1417.2\) seconds, or about \(23.6\) minutes, before irreversible body damage occurred.

Step by step solution

01

Compute the Temperature Change

The student needs to prevent the body temperature from rising to \(44^{\circ} \mathrm{C}\). The normal body temperature is \(37^{\circ} \mathrm{C}\). Therefore, we need to calculate the temperature change that occurs. The temperature change \(\Delta \mathrm{T}\) will be given by: \(\Delta \mathrm{T} = \mathrm{T}_{\mathrm{final}} - \mathrm{T}_{\mathrm{initial}} = 44 - 37 = 7^{\circ} \mathrm{C}\). It's given that \(1^{\circ} \mathrm{C} = 1 \mathrm{K}\), the temperature change \(\Delta \mathrm{T} = 7 \mathrm{K}\).
02

Calculate the Energy Required

The energy \(Q\) required to produce a change in temperature is given by: \(Q = mc\Delta \mathrm{T}\), where \(m\) is the mass of the body, \(c\) is the specific heat capacity and \(\Delta \mathrm{T}\) is the temperature change. Substituting the given values into this equation yields: \(Q = 70 \mathrm{~kg} \times 3480 \mathrm{~J} / \mathrm{kg.K} \times 7 \mathrm{K} = 1700640 \mathrm{~J}\).
03

Determine Time Required for Irreversible Damage

Given that the student is generating thermal energy at a rate of 1200 W, and \(1 \mathrm{W} = 1 \mathrm{J/s}\), we can then determine the time \(t\) it would take for the student's body to generate this amount of thermal energy. The time \(t\) is obtained by using the formula: \(t = Q / P\), where \(P\) is the power. Substituting the given values into the equation yields: \(t = 1700640 \mathrm{~J} / 1200 \mathrm{~J/s} = 1417.2 \mathrm{~s}\).

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Key Concepts

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

Specific Heat Capacity
The specific heat capacity of a substance is an important concept in thermodynamics. It is a measure of the amount of heat energy required to raise the temperature of a unit mass by one degree Celsius (or one Kelvin). In the context of the human body, the specific heat capacity tells us how much energy it takes to heat the body's mass. For a typical human body, the specific heat capacity is approximately 3480 J/kg·K.
This value is slightly lower than that of water, which is typically 4184 J/kg·K.
  • This difference occurs because body tissues contain proteins, fats, and minerals, all of which contribute lower specific heat capacities compared to water.
  • Understanding specific heat capacity helps in calculating how the body absorbs and distributes thermal energy.
Knowing the specific heat capacity is crucial, especially when considering physiological processes like body temperature regulation and how changes in body temperature can affect health and performance.
Thermal Energy Generation
Thermal energy generation refers to the production of heat within the body during physical activities such as running. This heat is primarily a byproduct of metabolic processes that convert food into energy, enabling muscles to work. As the problem indicates, our student generates thermal energy at a rate of 1200 watts during exercise.
This represents the thermal energy produced per second by the body.
  • Metabolic processes are inefficient, with a significant portion of energy being released as heat instead of motion.
  • The body needs to manage this heat generation efficiently to maintain a stable internal environment, which is essential for proper biological function.
In situations where heat is not properly released, such as when the cooling mechanisms of the body fail, this may lead to dangerous overheating conditions. Thus, understanding how much thermal energy the body generates aids in assessing how it manages or dissipates heat.
Temperature Change Calculation
Temperature change calculation is critical in assessing the impact of thermal energy on the body. To find out how much the body's temperature might rise, we use the formula: \[ \Delta T = T_{\text{final}} - T_{\text{initial}} \] In this exercise, the safe range of body temperature is up to 44°C; the normal body temperature is 37°C. Thus, the change in temperature, \( \Delta T \), is 7°C or 7K. This calculation assists in predicting how different rates of heat energy generation can affect body temperature.
  • It is crucial to consider such changes because even small alterations in temperature can impact enzyme function and protein stability.
  • Proteins may denature at elevated temperatures, leading to cellular damage and impaired bodily functions.
Therefore, predicting how temperature varies with changes in energy generation can help avert harmful physiological conditions.
Body Temperature Regulation
Body temperature regulation is a vital physiological function that maintains homeostasis. The human body employs several mechanisms to shed excess heat, such as sweating, vasodilation, and increasing blood circulation. These methods ensure that the body's core temperature remains around 37°C.
However, if these mechanisms fail, body temperature can rise dangerously.
  • Perspiration, for example, helps cool the body through evaporation.
  • Vasodilation expands blood vessels, allowing more blood to flow near the surface of the skin, where heat can disperse.
Despite these efficient systems, excessive thermal energy generation might exceed the body's ability to cool itself, potentially leading to overheating if conditions prevent heat loss. In the exercise, if the student were unable to dissipate the energy generated by running, the calculation shows the time until harmful level was reached is approximately 1417.2 seconds. Grasping the concepts of these regulatory processes ensures an understanding of how vital proper temperature control is to maintaining health and well-being.

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Most popular questions from this chapter

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

Animals in cold climates often depend on \(t w o\) layers of insulation: a layer of body fat (of thermal conductivity \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere \(1.5 \mathrm{~m}\) in diameter having a layer of fat \(4.0 \mathrm{~cm}\) thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at \(2.7^{\circ} \mathrm{C}\) during hibernation, while the inner surface of the fat layer is at \(31.0^{\circ} \mathrm{C}\). (a) What is the temperature at the fat-inner fur boundary so that the bear loses heat at a rate of \(50.0 \mathrm{~W} ?\) (b) How thick should the air layer (contained within the fur) be?

A \(\mathrm 500.0 \mathrm{~g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing \(1.00 \mathrm{~kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\). (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

An insulated beaker with negligible mass contains \(0.250 \mathrm{~kg}\) of water at \(75.0^{\circ} \mathrm{C}\). How many kilograms of ice at \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(40.0^{\circ} \mathrm{C}\) ?

A copper pot with a mass of \(0.500 \mathrm{~kg}\) contains \(0.170 \mathrm{~kg}\) of water, and both are at \(20.0^{\circ} \mathrm{C}\). A \(0.250 \mathrm{~kg}\) block of iron at \(85.0^{\circ} \mathrm{C}\) is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

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