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Chapter 1: Problem 65

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

An internally reversible refrigerator has a modified coefficient of performance accounting for realistic heat transfer processes of $$ \mathrm{COP}_{m}=\frac{q_{\text {in }}}{\dot{W}}=\frac{q_{\text {in }}}{q_{\text {out }}-q_{\text {in }}}=\frac{T_{c, i}}{T_{h, i}-T_{c, i}} $$ where \(q_{\text {in }}\) is the refrigerator cooling rate, \(q_{\text {out }}\) is the heat rejection rate, and \(\dot{W}\) is the power input. Show that \(\mathrm{COP}_{m}\) can be expressed in terms of the reservoir temperatures \(T_{c}\) and \(T_{h}\), the cold and hot thermal resistances \(R_{L, c}\) and \(R_{t, h}\), and \(q_{\text {in }}\), as $$ \mathrm{COP}_{m}=\frac{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}}{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{tot}}} $$ where \(R_{\mathrm{tot}}=R_{t, c}+R_{t, h}\). Also, show that the power input may be expressed as $$ \dot{W}=q_{\mathrm{in}} \frac{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{id \textrm {t }}}}{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}} $$

The diameter and surface emissivity of an electrically heated plate are \(D=300 \mathrm{~mm}\) and \(\varepsilon=0.80\), respectively. (a) Estimate the power needed to maintain a surface temperature of \(200^{\circ} \mathrm{C}\) in a room for which the air and the walls are at \(25^{\circ} \mathrm{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), may be approximated by an expression of the form \(h=0.80\left(T_{s}-T_{\infty}\right)^{1 / 3}\). (b) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.

A \(50 \mathrm{~mm} \times 45 \mathrm{~mm} \times 20 \mathrm{~mm}\) cell phone charger has a surface temperature of \(T_{s}=33^{\circ} \mathrm{C}\) when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity \(\varepsilon=0.92\) and is subject to a free convection heat transfer coefficient of \(h=4.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The room air and wall temperatures are \(T_{\infty}=22^{\circ} \mathrm{C}\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), respectively. If electricity costs \(C=\$ 0.18 / \mathrm{kW} \cdot \mathrm{h}\), determine the daily cost of leaving the charger plugged in when not in use.

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.

An inexpensive food and beverage container is fabricated from 25 -mm-thick polystyrene \((k=0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and has interior dimensions of \(0.8 \mathrm{~m} \times 0.6 \mathrm{~m} \times 0.6 \mathrm{~m}\). Under conditions for which an inner surface temperature of approximately \(2^{\circ} \mathrm{C}\) is maintained by an ice-water mixture and an outer surface temperature of \(20^{\circ} \mathrm{C}\) is maintained by the ambient, what is the heat flux through the container wall? Assuming negligible heat gain through the \(0.8 \mathrm{~m} \times\) \(0.6 \mathrm{~m}\) base of the cooler, what is the total heat load for the prescribed conditions?
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