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Chapter 13: Problem 2

at illustrates the concepts pertinent to this problem. A refrigerator has a surface area of \(5.3 \mathrm{~m}^{2}\). It is lined with \(0.075\) -m-thick insulation whose thermal conductivity is \(0.030 \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) The interior temperature is kept at \(5{ }^{\circ} \mathrm{C}\), while the temperature at the outside surface is \(25^{\circ} \mathrm{C}\). How much heat per second is being removed from the unit?

Short Answer

Expert verified
Heat removed is 42.4 J/s.

Step by step solution

01

Identify Known Values

Let's identify the given quantities in the problem: - Surface area, \( A = 5.3 \; \mathrm{m}^2 \)- Thickness of insulation, \( d = 0.075 \; \mathrm{m} \)- Thermal conductivity, \( k = 0.030 \; \mathrm{J/(s \cdot m \cdot 掳C)} \)- Interior temperature, \( T_{\text{inside}} = 5掳C \)- Exterior temperature, \( T_{\text{outside}} = 25掳C \)
02

Understand Heat Transfer Equation

The heat transfer through a material can be calculated using the formula:\[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]where \( Q \) is the heat transfer per second (watts), \( k \) is the thermal conductivity, \( A \) is the area, \( \Delta T \) is the temperature difference, and \( d \) is the thickness.
03

Calculate Temperature Difference

Determine the temperature difference between the interior and the exterior of the refrigerator:\[ \Delta T = T_{\text{outside}} - T_{\text{inside}} = 25掳C - 5掳C = 20掳C \]
04

Plug Values into Heat Transfer Equation

Using the heat transfer equation, substitute the known values:\[ Q = \frac{0.030 \cdot 5.3 \cdot 20}{0.075} \]
05

Compute the Heat Transfer Rate

Calculate the heat transfer rate by conducting the arithmetic operation:\[ Q = \frac{3.18}{0.075} = 42.4 \; \mathrm{J/s} \]
06

Conclusion

The amount of heat being removed from the refrigerator per second is 42.4 joules.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property of a material that tells us how well it can transfer heat. Imagine thermal conductivity as a highway for heat energy.
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When a material has high thermal conductivity, heat can move through it easily and quickly. Conversely, when the material has low thermal conductivity, heat moves slowly. This concept is crucial in designing devices like refrigerators.
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  • Definition: Thermal conductivity ( \( k \) ) is the measure of a material's ability to conduct heat.
  • Units: It is expressed in \( ext{J/(s} \cdot \text{m} \cdot \text{掳C)}\).
  • Material Dependency: Different materials have varying levels of conductivity, e.g., metals usually have high conductivity while insulating materials have low conductivity.
Understanding thermal conductivity helps us in selecting materials that correctly manage the flow of heat depending on whether we intend to retain or dissipate thermal energy.
Temperature Difference
Temperature difference, often symbolized as \( \Delta T \), represents the difference in temperature between two areas.
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This concept is analogous to pressure difference in fluids -- heat flows more rapidly when the temperature difference is greater.
.Let's explore some more about this important factor in heat transfer:
  • Formation: It is calculated as the difference between the temperature at two points. For example, between the inside and outside of a refrigerator.
  • Role in Heat Transfer: A larger temperature difference increases the heat transfer rate. This is evident in the heat transfer equation where \( \Delta T \) directly impacts \( Q \) (heat transfer rate).
  • Examples: Hot coffee cooling in a cold room or the warming effect when a cold drink is left in a warmer environment.
Recognizing the significance of temperature difference can help us understand and calculate how effective insulation is.
Insulation Thickness
Insulation thickness refers to how thick the insulating material is.
.
This thickness acts as a barrier, affecting how much heat flows through. Just like wearing a thicker jacket in winter keeps you warmer, thicker insulation prevents heat transfer better.
.The implications of insulation thickness in heat transfer are worth noting:
  • Functionality: The thickness ( \( d \)) of the insulating layer creates resistance against heat flow.
  • Heat Transfer Equation: In the equation \( Q = \frac{k \cdot A \cdot \Delta T}{d} \), an increase in \( d \) will reduce \( Q \), meaning less heat is transferred.
  • Material Considerations: The effectiveness also depends on the material's actual insulating properties, which is why both thickness and material type are critical.
Understanding and choosing the right insulation thickness is key to optimizing thermal efficiency and minimizing energy costs.

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Most popular questions from this chapter

A car parked in the sun absorbs energy at a rate of 560 watts per square meter of surface area. The car reaches a temperature at which it radiates energy at this same rate. Treating the car as a perfect radiator \((e=1)\), find the temperature.

A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders, each of which has the same length. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there?

In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissivity \(=0.75\) ). it has a temperature of \(62^{\circ} \mathrm{C}\). The new owner of the house paints the radiator a lighter color (emissivity \(=0.50\) ). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?

Two cylindrical rods are identical, except that one has a thermal conductivity \(k_{1}\) and the other has a thermal conductivity \(k_{2}\). As the drawing shows, they are placed between two walls that are maintained at different temperatures \(T_{\mathrm{W}}\) (warmer) and \(T_{\mathrm{C}}\) (cooler). When the rods are arranged as in part \(a\) of the drawing, a total heat \(Q^{\prime}\) flows from the warmer to the cooler wall, but when the rods are arranged as in part \(b,\) the total heat flow is \(Q\). Assuming that the conductivity \(k_{2}\) is twice as great as \(k_{1}\) and that heat flows only along the lengths of the rods, determine the ratio \(Q^{\prime} / Q\).

Two cylindrical rods have the same mass. One is made of silver (density \(=10\) \(\left.500 \mathrm{~kg} / \mathrm{m}^{3}\right)\), and one is made of iron (density \(\left.=7860 \mathrm{~kg} / \mathrm{m}^{3}\right)\). Both rods conduct the same amount of heat per second when the same temperature difference is maintained across their ends. What is the ratio (silver-to-iron) of (a) the lengths and (b) the radii of these rods?
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