Question

Calculate the amount of water (in liters) that would have to be vaporized at \(40^{\circ} \mathrm{C}\) (approximately body temperature) to expend the \(2.5 \times 10^{6}\) calories of heat generated by a person in one day (commonly called sweating). The heat of vaporization of water at this temperature is \(574 \mathrm{cal} / \mathrm{g}\). We normally do not sweat that much. What's wrong with this calculation? If \(1 \%\) of the energy produced as heat could be utilized as mechanical work, how large a weight could be lifted I meter?

Short answer

Vaporizing 4.356 liters of water is impractical. If 1% of energy is used, a 10660 kg weight can be lifted 1 meter.

Step-by-step solution

Identify the Known Values

From the problem, we know the following:- Total heat to be expended: \(2.5 \times 10^{6}\) calories- Heat of vaporization of water at \(40^{\circ} \mathrm{C}\): \(574 \mathrm{cal}/\mathrm{g}\)- Converting grams to liters (since the density of water is approximately \(1\mathrm{g/cm}^3\)), we need to find mass in grams.

Calculate the Mass of Water Needed

Use the formula for the heat required to vaporize water: \[ Q = m \cdot L_v \]where \(Q\) is the heat energy, \(m\) is the mass, and \(L_v\) is the heat of vaporization. Rearrange the formula to solve for mass \(m\):\[ m = \frac{Q}{L_v} \]Substitute the known values:\[ m = \frac{2.5 \times 10^{6} \text{ cal}}{574 \text{ cal/g}} \approx 4356 \text{ grams} \].

Convert Grams to Liters

Since the density of water is approximately 1 g/cm³, 4356 grams of water is roughly equivalent to 4356 cm³, or 4.356 liters.

Identify the Error in the Calculation

Sweating this much would mean vaporizing approximately 4.356 liters of water, which is not practical for the human body, as it far exceeds normal sweat production.

Calculate the Mechanical Work

If 1% of this energy could be utilized for mechanical work, calculate the amount of work possible:\[ \text{Mechanical work} = 0.01 \times 2.5 \times 10^{6} \text{ cal} = 25000 \text{ cal} \]Convert calories to Joules (1 cal = 4.184 J):\[ 25000 \text{ cal} \times 4.184 \text{ J/cal} = 104600 \text{ J} \].

Calculate the Weight Lifted

The work done \(W\) to lift a weight \(m\) a distance \(h\) against gravity \(g\) is: \[ W = m \cdot g \cdot h \]Rearrange to find mass (where \(g = 9.81 \text{ m/s}^2\), and \(h = 1 \text{ meter}\)):\[ m = \frac{W}{g \cdot h} = \frac{104600 \text{ J}}{9.81 \text{ m/s}^2 \times 1 \text{ m}} \approx 10660 \text{ kg} \].

Conclusion

A person would need to vaporize approximately 4.356 liters of water to dissipate the heat generated, which is unrealistic. If 1% energy were used for mechanical work, it could lift about 10660 kg by 1 meter.

Key concepts

Heat of Vaporization

In the realm of thermodynamics, the heat of vaporization is a crucial concept. It refers to the amount of heat required to convert a liquid, like water, into vapor without changing its temperature. During this phase transition, the substance absorbs energy to overcome intermolecular forces.
At body temperature, which is roughly 40°C, water's heat of vaporization is 574 cal/g. This value indicates that each gram of water needs 574 calories to become steam at this temperature.
Understanding this leads to the realization of how much energy the body would expend to evaporate large volumes of water, like in the sweating process. For example, the human body would theoretically need to vaporize approximately 4.356 liters of water to handle 2.5 million calories of heat, which showcases the immense heat transfer requirements and is why such a volume is impractical in biological systems. Sweating helps cool the body, but nowhere near these exaggerated amounts under normal conditions.

Mechanical Work in Biology

Mechanical work in biological systems hinges on converting chemical energy into kinetic energy. If we consider this conversion in the human body, only a fraction of total energy is utilized as mechanical work, given that most of it is lost as heat.
In the exercise provided, we see an interest in utilizing just 1% of total body heat for performing work. This translation from thermal energy to mechanical force is enlightening, showing that efficient energy use could potentially lift a weight. Specifically, if 1% of a person's daily heat production is harnessed, it amounts to roughly 25,000 calories or 104,600 Joules when converted.

• Assuming perfect efficiency, this can raise approximately 10,660 kg by 1 meter, which is a huge amount.
• In practical terms, while not feasible and highly simplified, such calculations offer insight into the body's capability to convert energy and perform creative work involving movement or resistance training.

Energy Conversion in Biological Processes

Energy conversion is vital in biological systems, ensuring organisms sustain various life processes. Our body constantly transforms energies; from caloric intake into mechanical work and thermal output.
In biology, the primary form of energy conversion faces many inefficiencies, especially when converting chemical energy from food into kinetic energy. Most of it gets dissipated as heat, illustrating significant energy loss in biological systems.

• Key examples of energy conversion include metabolic processes where nutrients generate ATP (adenosine triphosphate), which cells use as immediate energy source.
• Moreover, muscle contraction illustrates mechanical work from energy conversion – utilizing ATP, those muscles perform tasks.

The studied exercise highlights these inefficiencies by showcasing how minimal energy ends up in practical work form, reinforcing expectations within biological thermodynamics. Recognizing this helps understand the complex processes keeping organisms functional and proficient, though not fully effective in energy conservation terms.