Question

A person's head can be approximated as a 25-cm diameter sphere at \(35^{\circ} \mathrm{C}\) with an emissivity of \(0.95\). Heat is lost from the head to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(10^{\circ} \mathrm{C}\). Disregarding the neck, determine the total rate of heat loss from the head. (a) \(22 \mathrm{~W}\) (b) \(27 \mathrm{~W}\) (c) \(49 \mathrm{~W}\) (d) \(172 \mathrm{~W}\) (e) \(249 \mathrm{~W}\)

Short answer

Answer: (b) 27 W

Step-by-step solution

Calculate the surface area of the head.

First, we need to find the surface area of the head, which is approximated as a sphere. The formula for the surface area of a sphere is \(A = 4\pi r^2\), where \(r\) is the radius of the sphere. Since the diameter is given as 25 cm, we can find the radius by dividing the diameter by 2: \(r = \frac{25}{2} = 12.5 \, \mathrm{cm}\). Now, convert the radius to meters: \(r = 0.125 \, \mathrm{m}\). Calculate the surface area \(A\): \(A = 4\pi (0.125)^2 \approx 0.196 \, \mathrm{m}^2\).

Calculate the heat loss due to convection.

Now, we will determine the rate of heat loss due to convection. The formula for heat loss due to convection is \(Q_{conv} = hA(T_{head} - T_{air})\), where \(h\) is the heat transfer coefficient, \(A\) is the surface area, \(T_{head}\) is the temperature of the head, and \(T_{air}\) is the temperature of the surrounding air. Given the heat transfer coefficient \(h=11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(T_{head} = 35^{\circ} \mathrm{C}=308.15\mathrm{~K}\), and \(T_{air} = 25^{\circ} \mathrm{C}=298.15\mathrm{~K}\). Using these values, we can calculate the heat loss due to convection: \(Q_{conv} = 11 (0.196)(308.15-298.15) \approx 22 \, \mathrm{W}\).

Calculate the heat loss due to radiation.

Next, we will determine the rate of heat loss due to radiation. The formula for heat loss due to radiation is \(Q_{rad} = \epsilon \sigma A (T_{head}^4 - T_{surfaces}^4)\), where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \,\mathrm{W}/\mathrm{m}^2 \cdot \mathrm{K}^4)\), \(A\) is the surface area, \(T_{head}\) is the temperature of the head, and \(T_{surfaces}\) is the temperature of the surrounding surfaces. Given the emissivity \(\epsilon=0.95\), \(T_{head}=308.15\mathrm{~K}\), and \(T_{surfaces}=283.15\mathrm{~K}\) (since \(10^{\circ} \mathrm{C}=283.15\mathrm{~K}\)). Using these values, we can calculate the heat loss due to radiation: \(Q_{rad} = 0.95(5.67 \times 10^{-8})(0.196)(308.15^4 - 283.15^4) \approx 5 \, \mathrm{W}\)

Calculate the total heat loss.

Finally, we will add the heat loss due to convection and radiation to find the total heat loss: \(Q_{total} = Q_{conv} + Q_{rad} = 22 \, \mathrm{W} + 5 \, \mathrm{W} = 27 \, \mathrm{W}\) So, the total rate of heat loss from the person's head is \(27 \, \mathrm{W}\). The correct answer is (b) \(27 \, \mathrm{W}\).

Key concepts

Convection Heat Transfer

One of the primary ways that the human body loses heat to its surroundings is through convection heat transfer. This is the process by which heat is carried away by the movement of fluids, which in most cases involves the air surrounding the body. Imagine it like a gentle breeze that passes over your skin and removes some of the warmth.

The rate of convection heat loss can be calculated using the formula:
\[Q_{conv} = hA(T_{body} - T_{air})\]
In this equation, \(Q_{conv}\) represents the heat lost due to convection, \(h\) is the heat transfer coefficient that describes how well heat is transferred from the body to the air, \(A\) is the surface area of the body part losing heat, and \(T_{body} - T_{air}\) is the temperature difference between the body and the surrounding air.

For the body's head approximated as a sphere, this convection process can have significant impact on how warm you stay, especially on cold or windy days, where convection can increase and enhance heat loss.

Radiation Heat Transfer

Apart from the loss of heat through convection, our bodies also lose heat through radiation heat transfer. This type of heat transfer doesn't rely on any medium but occurs through the emission of infrared radiation. All objects, including the human body, emit some form of thermal radiation depending on their temperature.

The formula for this radiative heat loss is expressed as:
\[Q_{rad} = \[epsilon\] \[\sigma\] A (T_{body}^4 - T_{surroundings}^4)\]
Here, \(Q_{rad}\) is the heat lost through radiation, \(\epsilon\) is the emissivity of the body part (a measure of how effectively it emits heat radiation), \(\sigma\) represents the Stefan-Boltzmann constant, \(A\) stands for the surface area, and the difference in temperatures to the fourth power, \(T_{body}^4 - T_{surroundings}^4\), indicates how temperature differences affect radiant heat loss.

For humans, these radiative losses can be quite significant, and this is why we often feel colder when exposed to a cold environment even if the air is still.

Stefan-Boltzmann Constant

Central to the calculation of radiative heat transfer is the Stefan-Boltzmann constant, symbolized as \(\sigma\). It is a physical constant that plays a critical role in quantifying the total energy radiated per unit surface area of a black body per unit time.

The constant's value is approximately \(5.67 \times 10^{-8} \, \mathrm{W}/\mathrm{m}^2 \cdot \mathrm{K}^4\), and its significance in the formula for radiation heat transfer can't be overstated— it ensures that the power radiated by an object is proportional to the fourth power of its absolute temperature.

This factor is why radiative heat loss becomes more considerable for objects with higher temperatures, like the human body compared to the ambient surroundings— the warmer you are, the more heat you'll radiate away.

Spherical Surface Area Calculation

When assessing heat loss from a human head, we model it as a sphere to calculate the surface area, which is essential for both convection and radiation heat transfer calculations. The formula used to determine the surface area of any sphere is:
\[A = 4\pi r^2\]
where \(A\) symbolizes the area, \(\pi\) is the mathematical constant approximately equal to 3.14159, and \(r\) stands for the sphere's radius.

In the case of convection and radiation heat loss, the surface area is directly proportional to the amount of heat that can escape from the object. For understanding heat loss in biological terms, this calculation helps us to predict how changes in head size or shape can influence the body's thermal balance, making it a foundational concept in understanding thermal physiology and comfort.