An average human produces about \(10 \mathrm{M} \mathrm{J}\) of heat each day througn metabolic activity. If a human body were an isolated system of mass \(65 \mathrm{~kg}\) with the heat capacity of water, what temperature rise would the body experience? Human bodies are actually open systems, and the main mechanism of heat loss is through the evaporation of water. What mass of water should be evaporated each day to maintain constant temperature?

Short Answer

Expert verified
The temperature rise would be approximately \(2.39^\circ C\). To maintain constant temperature, about \(4.42 \mathrm{kg}\) of water would need to be evaporated each day.

Step by step solution

01

Calculate the Specific Heat Capacity of Water

To calculate the temperature rise, we need to know the specific heat capacity of water. It is a constant value, which is approximately \(4.186 \mathrm{J/g\cdot K}\) or \(4186 \mathrm{J/kg\cdot K}\). This is the amount of energy needed to raise the temperature of 1 kg of water by 1 degree Celsius.
02

Calculate the Temperature Rise

Use the formula for heat transfer, which is \(\text{heat} = \text{mass} \times \text{specific heat capacity} \times \text{change in temperature}\) or \(\text{Q} = \text{m} \times \text{c} \times \text{ΔT}\). Here \(\text{Q} = 10 \times 10^6 \mathrm{J}\), mass \(m = 65 \mathrm{kg}\), and specific heat capacity \(c = 4186 \mathrm{J/kg\cdot K}\). Rearrange the formula to solve for \(ΔT\) which gives \(\text{ΔT} = \text{Q} / (\text{m} \times \text{c})\). Substitute the values to calculate \(ΔT\).
03

Calculate the Evaporation Heat of Water

The heat of vaporization of water is approximately \(2260 \mathrm{kJ/kg}\), which is the amount of energy needed to turn 1 kg of water into vapor at boiling point. However, the evaporation occurs at body temperature, so the energy needed is slightly less but typically approximated to this value for such calculations.
04

Calculate the Mass of Water to be Evaporated

To maintain a constant temperature, the total heat produced by the body must be equal to the total heat lost by evaporation. Use the formula \(\text{heat lost} = \text{mass of water evaporated} \times \text{heat of vaporization}\) or \(\text{Q} = \text{m}_\text{water} \times \text{H}_\text{vap}\). Rearrange the formula to solve for the mass of water evaporated \(m_{\mathrm{water}}\) which gives \(m_{\mathrm{water}} = \text{Q} / \text{H}_\text{vap}\), and substitute the values to calculate the mass of water that needs to be evaporated.

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 Vaia!

Key Concepts

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

Specific Heat Capacity
Understanding the specific heat capacity is vital when studying how substances interact with heat. Specifically, the specific heat capacity of water is a key factor in many biological processes, including human physiology. It refers to the quantity of heat required to raise the temperature of one kilogram of water by one degree Celsius, which is approximately 4186 joules per kilogram per Kelvin (\(4186 \text{ J/kg} \times \text{K}\)). This high specific heat capacity means that water can absorb a lot of heat before its temperature rises significantly. This property is crucial for maintaining stable conditions within biological systems, as it allows for the absorption and release of heat without causing drastic temperature fluctuations that could be harmful to cellular processes.

Importance in Biological Systems

Water's high specific heat plays a role in temperature regulation in living organisms, and it also minimizes environmental temperature changes, providing a stable habitat for aquatic life.
Heat Transfer in Metabolism
Metabolism is the engine of life, a series of chemical reactions that provide the energy needed for all biological activities. Heat transfer in metabolism is an exchange of energy between the body and its surroundings. The heat produced is a byproduct of metabolic reactions, like breaking down food molecules for energy. In thermal biology, this interplay is critical since maintaining a constant body temperature is vital for the proper functioning of enzymatic reactions, and physiological processes.

The Role of Metabolic Heat

Metabolic heat contributes to maintaining body temperature but must be regulated because excessive heat can denature proteins and disrupt cellular structures. Humans, for instance, utilize various cooling methods, such as sweating and dilation of blood vessels, to dissipate excess heat and maintain homeostasis.
Evaporation Heat of Water
The evaporation heat, or heat of vaporization, of water is the energy needed to transform water from a liquid into a gas. This heat is vital for cooling mechanisms in biology, for example, when humans sweat. The approximate value of this heat is around 2260 kJ per kilogram at the boiling point of water. However, since evaporation in biological systems like our bodies occurs at lower temperatures, the actual energy required is somewhat less but is often approximated with this value to simplify calculations.

Cooling Effect Through Evaporation

When water evaporates from the skin, it absorbs this heat from the body, thereby reducing the body's overall temperature. This process is a part of the body's natural thermoregulation and allows for the dissipation of excess metabolic heat. It is especially effective because the evaporation heat of water is high, making it an efficient heat transfer method.
Energy Balance in Humans
The concept of energy balance in humans is about the equilibrium between energy intake and energy expenditure. Maintaining this balance is crucial for sustaining life and health. When we consume food, it's converted into energy through metabolic processes, and this energy is used for everything from cellular functions to physical activity.

Equilibrium Between Intake and Output

To maintain a constant body temperature, the heat produced from metabolic processes must be offset by heat loss, which can occur in various ways such as evaporation, radiation, convection, and conduction. When we exercise or experience elevated temperatures, for example, we perspire more to increase evaporation, which helps to dissipate heat. A failure to maintain energy balance can lead to overheating or hypothermia, depending on whether there is excessive heat production or inadequate heat loss, respectively. This balance is a key aspect of homeostasis, the body's ability to maintain a stable internal environment.

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

A gas obeys the equation of state \(V_{m}=R T / p+a T^{2}\) and its constantpressure heat capacity is given by \(C_{p^{\prime} m}=A+B T+C p\), where \(a, A, B\), and Care constants independent of \(T\) and \(p .\) Obtain expressions for (a) the JouleThomson coefficient and (b) its constant-volume heat capacity.

Silylene \(\left(\mathrm{SiH}_{2}\right)\) is a key intermediate in the thermal decomposition of silicon hydrides such as silane \(\left(\mathrm{SiH}_{4}\right)\) and disilane \(\left(\mathrm{Si}_{2} \mathrm{H}_{6}\right)\). Moffat et al. (H.K. Moffat, K.F. Jensen, and R.W. Carr, J. Phys. Chem. 95,145 (1991)) report \(\Delta_{1} H\) \(\left.\Theta_{\mathrm{SiH}_{2}}\right)=+274 \mathrm{kl} \mathrm{mol}^{-1} .\) If \(\Delta_{1} H^{\Theta}\left(\mathrm{SiH}_{4}\right)=+34.3 \mathrm{kj} \mathrm{mol}^{-1}\) and \(\Delta_{1} H^{\mathrm{O}}\left(\mathrm{Si}_{2}, \mathrm{H}_{6}\right)\) \(=+80.3 \mathrm{kj} \mathrm{mol}^{-1}\) (CRC Handbook (2004)), compute the standard enthalpies of the following reactions: a. \(\mathrm{SiH}_{4}(\mathrm{~g}) \rightarrow \mathrm{SiH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) b. \(\mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g}) \rightarrow \mathrm{SiH}_{2}(\mathrm{~g})+\mathrm{SiH}_{4}(\mathrm{~g})\)

Another alternative refrigerant (see preceding problem) is \(1,1,1,2\) tetrafluoroethane (refrigerant HFC-134a). Tillner-Roth and Baehr published a compendium of thermophysical properties of this substance (R. Tillner-Roth and H.D. Baehr, \(J\). Phys. Chem. Ref. Data 23, 657 (1994) from which properties such as the Joule-Thomson coefficient \(\mu\) can be computed. (a) Compute \(\mu\) at \(0.100 \mathrm{MPa}\) and \(300 \mathrm{~K}\) from the following data (all referring to \(300 \mathrm{~K}\) ): \(\begin{array}{llll} \text { P/MPa } & 0.080 & 0.100 & 0.12 \\ \text { Specific enthalpy/(k] } \mathrm{kg}^{-1} \text { ) } & 426.48 & 426.12 & 425.76 \end{array}\) (The specific constant-pressure heat capacity is \(\left.0.7649 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}\right)\) (b) Computer \(\mu\) at \(1.00 \mathrm{MPa}\) and \(350 \mathrm{~K}\) from the following data (all referring to \(350 \mathrm{~K}\) ): \(\begin{array}{llll} \text { PIMPa } & 0.80 & 1.00 & 1.2 \\ \text { Specific enthalpy/(k] } \mathrm{kg}^{-1} \text { ) } & 461.93 & 459.12 & 456.15 \end{array}\) (The specific constant-pressure heat capacity is \(1.0392 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}\).)

Since their discovery in 1985, fullerenes have received the attention of many chemical researchers. Kolesov et al. reported the standard enthalpy of combustion and of formation of crystalline \(\mathrm{C}_{60}\) based on calorimetric measurements (V.P. Kolesov, S.M. Pimenova, V.K. Pavlovich, N.B. Tamm, and A.A. Kurskaya, \(J\). Chem. Thermodynamics \(\mathbf{2 8}, 1121(1996)\) ). In one of their runs, they found the standard specific internal energy of combustion to be \(-36.0334 \mathrm{~kJ} \mathrm{~g}^{-1}\) at \(298.15 \mathrm{~K}\) Compute \(\Delta_{\mathrm{c}} H^{\Theta}\) and \(\Delta_{\mathrm{f}} H^{\Theta}\) of \(\mathrm{C}_{60}\)

Concerns over the harmful effects of chlorofluorocarbons on stratospheric ozone have motivated a search for new refrigerants. One such alternative is 2,2 -dichloro-1,1,1-trifluoroethane (refrigerant 123). Younglove and McLinden published a compendium of thermophysical properties of this substance (B.A. Younglove and M. McLinden, \(J\). Phys. Chem. Ref Data 23,7 (1994)), from which properties such as the Joule-Thomson coetficient \(\mu\) can be computed. (a) Compute \(\mu\) at \(1.00\) bar and \(50^{\circ} \mathrm{C}\) given that \((\partial H / \partial p)_{T},=-3.29\) \(\times 10^{3} \mathrm{~J} \mathrm{MPa}^{-1} \mathrm{~mol}^{-1}\) and \(C_{p \cdot m}=110.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\). (b) Compute the temperature change that would accompany adiabatic expansion of \(2.0 \mathrm{~mol}\) of this refrigerant from \(1.5\) bar to \(0.5\) bar at \(50^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free