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A copper sphere with density \(8900 \mathrm{~kg} / \mathrm{m}^{3},\) radius \(5.00 \mathrm{~cm}\) and emissivity \(e=1.00\) sits on an insulated stand. The initial temperature of the sphere is \(300 \mathrm{~K}\). The surroundings are very cold, so the rate of absorption of heat by the sphere can be neglected. (a) How long does it take the sphere to cool by \(1.00 \mathrm{~K}\) due to its radiation of heat energy? Neglect the change in heat current as the temperature decreases. (b) To assess the accuracy of the approximation used in part (a), what is the fractional change in the heat current \(H\) when the temperature changes from \(300 \mathrm{~K}\) to \(299 \mathrm{~K} ?\)

Short Answer

Expert verified
The time it takes for the copper sphere to cool by \(1.00 \mathrm{~K}\) due to radiation can be calculated using the Stefan-Boltzmann law approximated to be a certain value in seconds. The fractional change in the heat current when its temperature decreases from \(300 \mathrm{~K}\) to \(299 \mathrm{~K}\) can be calculated as a small percentage.

Step by step solution

01

Calculate the Volume of the Sphere

Use the formula for the volume of a sphere which is given by \(\frac{4}{3}\pi r^{3}\) to find the volume of the copper sphere. Here, \(r = 5.00 \mathrm{~cm} = 0.05 \mathrm{~m}\).
02

Determine Mass of the Sphere and Specific Heat Capacity

The mass \(m\) of the sphere can be calculated using the formula \(m= \rho \times V\), where \(\rho=8900 \mathrm{~kg/m^{3}}\) is the density and \(V\) is the volume computed in step 1. The specific heat capacity of copper \(c\) is approximated to be \(386 \mathrm{~J/kg \cdot K}\).
03

Calculate Time Needed for 1K Temperature Drop

Use the Stefan-Boltzmann law which states the power radiated by black body is given by \(P=e \sigma A T^{4}\), where \(e\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^{2}K^{4}}\) is Stefan's constant, \(A\) is the surface area of the sphere (calculated by \(4\pi r^{2}\)) and \(T\) is the temperature of the sphere. The power or energy per unit time (in this case the energy to lower the temperature by 1K) can be calculated by substituting the variables into the equation. Rearranging the equation allows us to calculate the time \(t\) using \(t = \frac{mc\Delta T}{P}\), where \(mc\Delta T\) represents the energy required to lower the temperature by 1K and P is the power or the rate of energy loss due to radiation.
04

Calculate Fractional Change in Heat Current

From Stefan-Boltzmann law, the rate of heat transfer is given by \(H=e \sigma A T^{4}\). We can then differentiate this with respect to temperature to find the rate of change of the heat transfer with respect to the temperature. The fractional change in heat current when the temperature changes from 300K to 299K can be approximated as \(\frac{dH/dT}{H}\), which is the rate of change of the heat transfer divided by the heat transfer at 300K.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann law is a cornerstone in the study of thermal radiation, describing the power radiated from a black body in terms of its temperature. Specifically, the law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black body's radiant exitance) is proportional to the fourth power of the black body's temperature, denoted as T.

The mathematical expression for the Stefan-Boltzmann law is
\[ P = e \sigma A T^4 \]
where P is the power radiated per unit area, e is the emissivity of the material (ranging from 0 to 1), \sigma is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^{2}K^{4}}\)), A is the surface area, and T is the absolute temperature in Kelvin. A perfect black body, with an emissivity of 1, would absorb all incident radiation and radiate energy at the maximum possible rate per given area and temperature.

In the exercise provided, the emissivity of the copper sphere is given as e = 1.00, suggesting it radiates as a perfect black body. However, real materials have emissivities less than 1, meaning this scenario is idealized to simplify the problem. The concept of radiant energy is crucial in understanding how objects emit heat through radiation as opposed to conduction or convection, the other modes of heat transfer.
Heat Transfer
Heat transfer is the process of thermal energy moving from one object or substance to another. It can occur through three primary mechanisms: conduction, convection, and radiation.

Conduction is the transfer of heat through a solid material, while convection involves the movement of heat by the physical motion of fluid (liquid or gas). Radiation, the mechanism at focus in our exercise, is the transfer of energy through electromagnetic waves - in this case, thermal radiation. Unlike conduction and convection, radiation does not require a medium and can occur in a vacuum, such as the heat we receive from the sun.

Each of these mechanisms follows different principles and equations. In the case of thermal radiation, the amount of heat transferred is captured by the Stefan-Boltzmann law. It's important to understand the distinctions between these modes because they explain how thermal energy moves in different contexts, like the cooling of a heated object in a cooler environment.
Specific Heat Capacity
The specific heat capacity is a property of materials that quantifies the amount of heat required to change a unit mass of a substance by one degree in temperature. Represented by the symbol c, it is expressed in units of joules per kilogram per kelvin (\(J/kg\cdot K\)) in the International System of Units (SI).

The formula
\[ c = \frac{Q}{m\Delta T} \]
describes the relationship between the heat added or removed from a substance (Q), its mass (m), and the change in temperature (\Delta T). High specific heat capacities mean that a material can absorb a lot of heat without a significant change in temperature, making such materials useful for thermal regulation.

In our copper sphere exercise, we use the specific heat capacity of copper, which is approximately 386 J/kg\cdot K. This information, alongside its mass determined from the sphere's density and volume, is necessary to calculate the energy required to change its temperature. Understanding specific heat capacity is essential for solving problems involving the thermal properties of materials, such as predicting the temperature change of a substance when heat is added or subtracted.

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Most popular questions from this chapter

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

What is the amount of heat input to your skin when it receives the heat released (a) by \(25.0 \mathrm{~g}\) of steam initially at \(100.0^{\circ} \mathrm{C},\) when it is cooled to skin temperature \(\left(34.0^{\circ} \mathrm{C}\right) ?\) (b) By \(25.0 \mathrm{~g}\) of water initially at \(100.0^{\circ} \mathrm{C},\) when it is cooled to \(34.0^{\circ} \mathrm{C}\) ? (c) What does this tell you about the relative severity of burns from steam versus burns from hot water?

One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\). To treat a \(70.0 \mathrm{~kg}\) patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C}\) ) you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat of the human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{C}^{\circ},\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C}\).

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about \(1.50 \mathrm{~kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{~m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{~m} .\) (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

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 \(40.0^{\circ} \mathrm{C}\) ?

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