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Hot coffee is contained in a cylindrical thermos bottle that is of length \(L=0.3 \mathrm{~m}\) and is lying on its side (horizontally). The coffee containet consists of a glass flask of diameter \(D_{1}=0.07 \mathrm{~m}\), separated from an aluminum housing of diameter \(D_{2}=0.08 \mathrm{~m}\) by air at atmospheric pressure. The outer surface of the flask and the inner surface of the housing are silver coated to provide emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.25\). If these sarface temperatures are \(T_{1}=75^{\circ} \mathrm{C}\) and \(T_{2}=35^{\circ} \mathrm{C}\). what is the heat loss froen the coffee?

Short Answer

Expert verified
The heat loss from the coffee in the thermos bottle is approximately 8.59 W.

Step by step solution

01

Calculate the Surface Area of the Flask and Housing

Since both the glass flask and the aluminum housing are cylindrical, we first need to find their surface areas where the heat loss occurs. The outer surface area of the glass flask and inner surface area of the aluminum housing can be calculated using the formula for the lateral surface area of a cylinder: \(A = 2\pi rL\) Where \(A\) is the surface area, \(r\) is the radius of the cylinder, and \(L\) is the length of the cylinder. Now, calculate the surface area of the flask: \(A_{1} = 2\pi \frac{D_{1}}{2}L\) \(A_{1} = 2\pi \times \frac{0.07\,\text{m}}{2} \times 0.3\,\text{m}\)
02

Calculate the Emissivity-Area Product

Next, we will calculate the emissivity-area (EA) product for the two cylindrical surfaces. The EA product represents the heat transfer between the two surfaces due to radiation: \(\text{EA} = \varepsilon_{1} A_{1} = \varepsilon_{2} A_{2}\) Since \(\varepsilon_{1} = \varepsilon_{2}\), the surface area of the flask can be used for calculating the EA product: \(\text{EA} = 0.25 A_{1}\)
03

Calculate the Heat Loss due to Radiation

Now, we can calculate the heat transfer between the two surfaces due to radiation, using the Stefan-Boltzmann Law for two surfaces at different temperatures: \(Q = \frac{\text{EA} \sigma (T_{1}^{4} - T_{2}^{4})}{D_{1}}\) Where: - \(Q\) is the heat transfer between the surfaces due to radiation (W) - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times 10^{-8} \,\text{W m}^{-2} \text{K}^{-4}\)) - \(T_{1}\) and \(T_{2}\) are the surface temperatures of the flask and housing, respectively (K) First, convert given temperatures from °C to K: \(T_{1} = 75^{\circ}\text{C}+273.15 = 348.15\,\text{K}\) \(T_{2} = 35^{\circ}\text{C}+273.15 = 308.15\,\text{K}\) Next, substitute the given temperature values and the EA product into the equation to find the heat loss due to radiation: \(Q = \frac{(0.25 A_{1}) \times 5.67 \times 10^{-8} \times (348.15^{4}-308.15^{4})}{0.07}\)
04

Find Heat Loss from the Coffee

The heat loss from the coffee will be the same as the heat loss due to radiation calculated above. Calculate the value using the given values and equations: \(Q = \frac{(0.25 \times 2\pi \times \frac{0.07}{2} \times 0.3) \times 5.67 \times 10^{-8} \times (348.15^{4}-308.15^{4})}{0.07}\) \(Q = 8.59\,\text{W}\) Therefore, the heat loss from the coffee is approximately 8.59 W.

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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 fundamental principle in understanding how thermal radiation leads to heat transfer between objects. It states that the power radiated per unit area from a black body is directly proportional to the fourth power of its absolute temperature. The formula for this relationship is given by:

\[ P = \frac{Q}{A} = \text{Stefan-Boltzmann constant}\times \text{emissivity} \times T^{4} \]

The constant involved is the Stefan-Boltzmann constant (\( \text{Stefan-Boltzmann constant} \)), and it has a value of approximately \( 5.67\times10^{-8} \text{W/m}^2\text{K}^{-4} \). Emissivity is a dimensionless factor that measures how effectively a real surface emits thermal radiation compared to an ideal black body, which has an emissivity of 1. Temperature (\( T \)) must be in Kelvins for the law to be applied accurately. This law is particularly helpful in predicting the thermal radiation heat transfer in many applications, such as cooling electronic equipment, estimating Earth's radiation balance, and, as seen in the example, calculating the heat loss from a thermos bottle.
Thermal Radiation
Thermal radiation refers to the emission of electromagnetic waves from the surface of an object, which occurs due to the object's temperature. It is a form of heat transfer that does not require any medium to travel, which means it can even occur through the vacuum of space.

All objects with a temperature above absolute zero emit thermal radiation. The amount and type of radiation emitted depend on the object's temperature and surface properties. For instance, high-temperature surfaces emit more radiation and at shorter wavelengths compared to cooler surfaces. Unlike conduction and convection, which require physical contact or a fluid medium, thermal radiation can affect an object's temperature from a distance. This type of heat transfer is significant in various real-life scenarios, including our thermos bottle example and even in astrophysics, where it helps us understand the heat transfer between stars and planets.
Cylindrical Surface Area
The surface area of a cylinder is an essential factor when analyzing heat transfer from cylindrical objects like pipes, tanks, or in this case, a coffee thermos. The total surface area of a cylinder includes the area of two circles (the bases) and the area of the rectangle that forms the curved surface, known as the lateral surface area.

The formula to calculate the lateral surface area (\( A \)) of a cylinder is given by:

\[ A = 2\text{pi} \times r \times L \]

where \( r \) is the radius of the cylinder, \( L \) is the height or length, and \( \text{pi} \) is the mathematical constant approximately equal to 3.14159. In heat transfer problems involving cylinders, understanding how to calculate this area is crucial because it directly influences the amount of heat exchanged through the surface.
Emissivity
Emissivity is a measurement of an object's ability to emit infrared energy compared to a perfect thermal radiator, known as a black body. Emissivity is dimensionless and ranges from 0 to 1, with 1 being a perfect emitter and 0 reflecting no thermal radiation.

Real surfaces do not behave exactly like ideal black bodies; they have emissivities less than 1, meaning they emit less thermal radiation than a black body at the same temperature. The emissivity value depends on factors such as the material's properties, its surface finish, and the wavelength of emitted radiation. In the thermos bottle example, the silver coating had an emissivity of 0.25, acknowledging a reduced emission capability compared to a perfect emitter. Accounting for emissivity is crucial in accurately calculating thermal radiation heat transfer, including in complex systems such as thermal insulation in buildings or the thermal management of electronic devices.

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Most popular questions from this chapter

A flat-bottomed hole \(6 \mathrm{~mm}\) in diameter is bored to a depth of \(24 \mathrm{~mm}\) in a diffuse, gray material having an emissivity of \(0.8\) and a uniform temperature of \(1000 \mathrm{~K} .\) (a) Determine the radiant power leaving the opening of the cavity. (b) The effective emissivity \(\varepsilon_{e}\) of a cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening and a temperature of the inner surfaces of the cavity. Calculate the effective emissivity of the cavity described above. (c) If the depth of the hole were increased, would \(\varepsilon_{e}\) increase or decrease? What is the limit of \(s_{\epsilon}\) as the depth increases?

A wall-mounted natural gas heater uses combustion on a porous catalytic pad to maintain a ceramic plate of emissivity \(\varepsilon_{c}=0.95\) at a uniform temperature of \(T_{c}=1000 \mathrm{~K}\). The ceramic plate is separated from a glass plate by an air gap of thickness \(L=50 \mathrm{~mm}\). The surface of the glass is diffuse, and its spectral transmissivity and absorptivity may be approximated as \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=1\) for \(0 \leq \lambda \leq 0.4 \mu \mathrm{m}, \tau_{\lambda}=1\) and \(\alpha_{\lambda}=0\) for \(0.4<\lambda \leq 1.6 \mu \mathrm{m}\), and \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=0.9\) for \(\lambda>1.6 \mu \mathrm{m}\). The exterior sarface of the glass is exposed to quiescent ambient air and large surroundings for which \(T_{m}=T_{a x}=300 \mathrm{~K}\). The height and width of the heater are \(H=W=2 \mathrm{~m}\). (a) What is the total transmissjvity of the glass to irradiation from the ceramic plate? Can the glass be approximated as opaque and gray? (b) For the prescribed conditions, evaluate the glass temperature, \(T_{R}\), and the rate of heat transfer from the heater, \(q_{k}\). (c) A fan may be used to control the convection coefficient \(h_{n}\) at the exterior surface of the glass. Compute and plot \(T_{g}\) and \(q_{\mathrm{b}}\) as a function of \(h_{e}\) for \(10 \leq h_{0} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A radiant heater, which is used for surface treatment processes, consists of a long cylindrical heating element of diameter \(D_{1}=0.005 \mathrm{~m}\) and emissivity \(\varepsilon_{1}=0.80\). The heater is partially enveloped by a long. thin parabolic reflector whose inner and outer surface emissivities are \(\varepsilon_{24}=0.10\) and \(\varepsilon_{20}=0.80\), respectively. Inner and outer surface areas per unit length of the reflector are each \(A_{2}^{\prime}=A_{20}^{\prime}=0.20 \mathrm{~m}\), and the average convection coefficient for the combined inner and outer surfaces is \(\bar{h}_{2 \dot{m}}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The system may be assumed to be in an infinite, quiescent medium of atmospheric air at \(T_{-}=300 \mathrm{~K}\) and to be exposed to large surroundings at \(T_{\text {aur }}=300 \mathrm{~K}\). (a) Sketch the appropriate radiation circuit, and write expressions for each of the network resistances. (b) If, under steady-state conditions, electrical power is dissipated in the heater at \(P_{1}^{\prime}=\) \(1500 \mathrm{~W} / \mathrm{m}\) and the heater surface temperature is \(T_{1}=1200 \mathrm{~K}\), what is the net rate at which radiant energy is transferred from the heater? (c) What is the net rate at which radiant energy is transferred from the heater to the surroundings? (d) What is the temperature, \(T_{2}\), of the reflector?

A radiometer views a small target (1) that is being heated by a ring-shaped disk heater (2). The target has an area of \(A_{1}=0.0004 \mathrm{~m}^{2}\), a temperature of \(T_{1}=\) \(500 \mathrm{~K}\), and a diffuse, gray emissivity of \(s_{1}=0.8\). The heater operates at \(T_{2}=1000 \mathrm{~K}\) and has a black surface. The radiometer views the entire sample area with a solid angle of \(\omega=0.0008 \mathrm{sr}\). (a) Write an expression for the radiant power leaving the target which is collected by the radiometer, in terms of the target radiosity \(J_{1}\) and relevant geometric parameters. Leave in symbolic form. (b) Write an expression for the target radiosity \(J_{1}\) in terms of its irradiation, emissive power, and appropriate radiative properties. Leave in symbolic form. (c) Write an expression for the irradiation on the target, \(G_{1}\), due to emission from the heater in terms of the heater emissive power, the heater area, and an appropriate view factor. Use this expression to numerically evaluate \(G_{1}\). (d) Use the foregoing expressions and results to determine the radiant power collected by the radiometer.

A laboratory oven has a cubical interior chamber \(1 \mathrm{~m}\) on a säde with interior sarfaces that are of emissivity \(z-0.85\). Determine the initial rate of radiation heat transfer to the laboratory in which the oven is placed when the oven door is opened. The oven and surroundings temperatures are \(T_{w}=375^{\circ} \mathrm{C}\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), respectively. Treat the oven door as one surface and the remaining five interior furnace walls as another.

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