Question

A student wants to lose some weight. He knows that rigorous aerobic activity uses about \(700 \mathrm{kcal} / \mathrm{h}(2900 \mathrm{kJ} / \mathrm{h})\) and that it takes about 2000 kcal per day \((8400 \mathrm{kJ})\) just to support necessary biological functions, including keeping the body warm. He decides to burn calories faster simply by sitting naked in a \(16^{\circ} \mathrm{C}\) room and letting his body radiate calories away. His body has a surface area of about \(1.7 \mathrm{m}^{2}\) and his skin temperature is \(35^{\circ} \mathrm{C} .\) Assuming an emissivity of \(1.0,\) at what rate (in \(\mathrm{kcal} / \mathrm{h}\) ) will this student "burn" calories?

Short answer

The student burns approximately 169.42 kcal/h.

Step-by-step solution

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law describes the power radiated from a black body in terms of its temperature. The formula is given by \( P = \epsilon \sigma A ( T^4 - T_c^4 ) \), where \( P \) is the power radiated, \( \epsilon \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} \mathrm{W/m^2 K^4} \)), \( A \) is the surface area, \( T \) is the body's absolute temperature in Kelvin, and \( T_c \) is the surrounding absolute temperature in Kelvin.

Convert Temperatures to Kelvin

The body's temperature is given in Celsius as \(35^{\circ} \mathrm{C}\), and the room's temperature as \(16^{\circ} \mathrm{C}\). Convert these to Kelvin by adding 273.15: \( T = 35 + 273.15 = 308.15 \mathrm{K} \) and \( T_c = 16 + 273.15 = 289.15 \mathrm{K} \).

Calculate the Power Radiated

Substitute the values into the Stefan-Boltzmann Law. With \( \epsilon = 1.0 \), \( \sigma = 5.67 \times 10^{-8} \mathrm{W/m^2 K^4} \), \( A = 1.7 \mathrm{m}^2 \), \( T = 308.15 \mathrm{K} \), and \( T_c = 289.15 \mathrm{K} \), we have:\[P = 1.0 \times 5.67 \times 10^{-8} \times 1.7 \times (308.15^4 - 289.15^4) \].

Simplify and Solve for Power

Calculate the powers of the temperatures:\(308.15^4 = 9.028 \times 10^9 \) and \(289.15^4 = 6.976 \times 10^9 \). Therefore,\( P = 1.0 \times 5.67 \times 10^{-8} \times 1.7 \times (9.028 \times 10^9 - 6.976 \times 10^9) \).Simplifying gives \( P = 1.0 \times 5.67 \times 10^{-8} \times 1.7 \times 2.052 \times 10^9 \), which results in \( P \approx 197 \mathrm{W} \).

Convert Power to Energy per Hour in kcal

Convert using the relationship \(1 \, \mathrm{W} = 0.860 \, \mathrm{kcal/h}\):\( 197 \, \mathrm{W} \times 0.860 \, \mathrm{kcal/h/W} = 169.42 \, \mathrm{kcal/h} \).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a principle in thermodynamics that helps us understand how objects radiate energy. It states that the power emitted by a "black body"—an idealized physical body that absorbs all falling radiation—depends on its temperature. This is mathematically expressed by the formula \( P = \epsilon \sigma A ( T^4 - T_c^4 ) \). Here, \( P \) represents the radiant energy power, \( \epsilon \) is the emissivity of the body (with 1 representing a perfect emitter), \( \sigma \) is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} \mathrm{W/m^2 K^4} \)), \( A \) is the surface area, \( T \) is the object's temperature in Kelvin, and \( T_c \) is the surrounding environment's temperature in Kelvin.

For the human body, understanding this law can reveal how it emits heat energy, as our bodies radiate thermal energy much like a black body does. By applying this law, one can calculate how much energy the body releases into its surroundings.

Heat Radiation

Heat radiation involves the transfer of energy in the form of electromagnetic waves or photons. All objects emit energy in this manner because of their temperature. Typically, the warmer the object, the more energy it radiates. When you sit in a cooler room, your body loses heat energy to the environment through such radiation.

This kind of energy transfer doesn't need any medium. It enables the body to lose heat even in space, where other forms of heat transfer like conduction or convection are impossible. For practical uses, understanding heat radiation is crucial for managing temperature in biological and non-biological systems.

Energy Conversion

Energy conversion, in this context, refers to transforming stored bodily energy into heat that is radiated away. When considering the human body's energy usage, basal metabolic rate plays a significant role. This rate represents the amount of energy required to maintain basic bodily functions such as breathing, circulation, and cellular processes.

In this scenario, the body converts chemical energy stored in calories into heat energy, which is then radiated away. By monitoring and calculating such energy conversions, one can estimate how physical processes like staying in a cold room can contribute to burning calories, integrating physics principles into biological contexts.

Biological Functions

Biological functions pertain to the processes and activities needed to maintain life. These include vital activities such as maintaining a stable body temperature, breathing, and circulating blood. Your body constantly undergoes chemical reactions to sustain these functions, utilizing energy derived from food.

In the exercise, the student is interested in losing weight through thermal radiation by opting to sit in a cooler room. This method leverages the principle of achieving energy balance. By increasing the difference between the body's internal and external temperatures, one can heighten energy expenditure through radiation, effectively "burning calories," which helps in weight management.