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A freezer compartment is covered with a 2 -mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

Short Answer

Expert verified
The time required to completely melt the frost is approximately 11,690 seconds (about 3.25 hours).

Step by step solution

01

Understand the given data

We are given the following information: - Initial thickness of the frost (d) = 2mm - Ambient air temperature (T_a) = 20掳C - Heat transfer coefficient (h) = 2 W/m虏K - Mass density of the frost (蟻) = 700 kg/m鲁 - Latent heat of fusion of the frost (L_f) = 334 kJ/kg
02

Calculate the temperature difference

We know that the frost is at 0掳C (since it's at the melting point) and the ambient air is at 20掳C. The temperature difference (螖T) can be calculated as: 螖T = T_a - T_f 螖T = 20掳C - 0掳C = 20掳C
03

Calculate the heat transfer rate

Using the heat transfer coefficient (h) and temperature difference (螖T), we can estimate the heat transfer rate (Q) per unit area of the frost: Q = h * 螖T Q = 2 W/m虏K * 20 K = 40 W/m虏
04

Calculate the volume of the frost per unit area

Next, we can calculate the volume of the frost per unit area (V) by dividing the initial thickness (d) by 1000 to change the units from mm to m, and using volume = density * mass: V = (2 mm / 1000) = 0.002 m鲁/m虏
05

Calculate the mass of the frost per unit area

To find the mass of the frost per unit area, we can use the mass density (蟻) and volume (V) calculated in the previous step: mass_per_unit_area = 蟻 * V mass_per_unit_area = 700 kg/m鲁 * 0.002 m鲁/m虏 = 1.4 kg/m虏
06

Calculate the total heat required to melt the frost

Now, we can calculate the total amount of heat (Q_total) required to melt the frost per unit area by multiplying the mass per unit area with the latent heat of fusion (L_f): Q_total = mass_per_unit_area * L_f Q_total = 1.4 kg/m虏 * 334 kJ/kg = 467.6 kJ/m虏
07

Estimate the time required to melt the frost

Converting the heat transfer rate (Q) from W/m虏 to kJ/m虏: Q = 40 W/m虏 * (1 kJ / 1000 W) = 0.04 kJ/m虏s Finally, we can estimate the time (t) required to melt the frost per unit area by dividing the total heat required (Q_total) with the heat transfer rate (Q): t = Q_total / Q t = 467.6 kJ/m虏 / 0.04 kJ/m虏s = 11690 s Therefore, it will take approximately 11,690 seconds (about 3.25 hours) to completely melt the frost.

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Key Concepts

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

Natural Convection
Natural convection is a vital concept in heat transfer, where fluid motion results from differences in density due to temperature gradients. This process does not require any external device, like a fan or pump, to move the fluid. Instead, nature takes care of this moving through buoyancy effects.

In this exercise, the freezer's exposed surface loses heat through natural convection. The warmer ambient air at 20掳C prompts the transfer of thermal energy to the colder layer of frost. The crucial factor here is the heat transfer coefficient (\(h = 2 \mathrm{ \ W/m}^2K\)) which measures how efficiently heat is transferred during natural processes. This coefficient depends on various factors, including surface properties and the temperature difference.

Understanding how natural convection works gives insight into why and how objects cool or heat naturally, influencing everything from weather systems to heating in homes.
Latent Heat of Fusion
Latent heat of fusion is a specific amount of energy needed to convert a solid into a liquid at its melting point, without changing the temperature. This concept is crucial for understanding how substances like ice melt.

In our problem, the latent heat of fusion, given as 334 kJ/kg, represents the energy required to melt the frost completely. Each kilogram of frost needs this amount of energy to change its state from solid to liquid. It helps explain why even when the freezer compartment's temperature stops dropping, the frost doesn't immediately turn to water鈥攊t first absorbs energy equal to its latent heat of fusion.

The phenomenon is important in numerous practical applications, including refrigeration and climate science, where phase changes play a significant role.
Mass Density
Mass density is a measure of mass per unit volume, expressed typically in kilograms per cubic meter (\(kg/m^3\)). It is a key property in dealing with materials and comparing them.

For the frost in this exercise, the density is 700 kg/m鲁. This helps determine how much mass is present in a given space and how much energy is involved in processes such as melting.The frost's mass density enables us to convert volume into mass. Using this, we determine the total mass of frost covering the freezer's surface.

Understanding density is critical in physics as it impacts how substances interact, float, sink, or require heat for phase changes. Land and water breezes, engine efficiency, and even baking rely on density.
Temperature Difference
The temperature difference, denoted usually as \(\Delta T\), plays a central role in driving heat transfer. In the most basic terms, it refers to the difference between two temperatures.In our context, the temperature difference is between the ambient air at 20掳C and the frost at 0掳C, calculated as 20掳C. This difference determines how fast or slow heat transfer occurs between two areas.

Bigger temperature differences result in faster heat transfer rates, and thus influence how quickly the frost melts. Thus, \(\Delta T\) is integral for applying Newton's Law of Cooling in determining the heat transfer rate.Temperature differences are universally essential, impacting every heat-related process, from cooking to climate control, and in engineering systems everywhere.

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Most popular questions from this chapter

What is the thickness required of a masonry wall having thermal conductivity \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) if the heat rate is to be \(80 \%\) of the heat rate through a composite structural wall having a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) and a thickness of \(100 \mathrm{~mm}\) ? Both walls are subjected to the same surface temperature difference.

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the roof under steadystate conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the same ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airflow conditions over the roof, increasing with increasing air speed. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

An overhead 25-m-long, uninsulated industrial steam pipe of \(100-\mathrm{mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{\circ} \mathrm{C}\), and the coefficient associated with natural convection is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steam is generated in a gas-fired boiler operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02\) per \(\mathrm{MJ}\), what is the annual cost of heat loss from the line?

The heat flux that is applied to the left face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The wall is of thickness \(L=10\) \(\mathrm{mm}\) and of thermal conductivity \(k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the surface temperatures of the wall are measured to be \(50^{\circ} \mathrm{C}\) on the left side and \(30^{\circ} \mathrm{C}\) on the right side, do steady-state conditions exist?

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