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The \(R\) factor for housing insulation gives the thermal resistance in units of \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) h/BTU. A good wall for harsh climates, corresponding to about 10.0 in of fiberglass, has \(R=40.0 \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\) a) Determine the thermal resistance in SI units. b) Find the heat flow per square meter through a wall that has insulation with an \(R\) factor of \(40.0,\) when the outside temperature is \(-22.0^{\circ} \mathrm{C}\) and the inside temperature is \(23.0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The heat flow per square meter through the wall is approximately 6.39 W/m².

Step by step solution

01

Convert R factor to SI units

To convert \(R\) from \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) h/BTU to SI units (m² K/W), we need to use the following conversion factors: 1 ft = 0.3048 m 1 °F = 5/9 K 1 h = 3600 s 1 BTU = 1055.06 J So, the conversion factor for \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) h/BTU to m² K/W is: \(\frac{0.3048^2 \cdot 5/9 \cdot 3600}{1055.06} = 0.1761\) Now, multiply this conversion factor by the given \(R\) value (40.0 ft² °F h/BTU): \(R_{SI} = 40 \cdot 0.1761 = 7.044\) m² K/W
02

Calculate the heat flow per square meter

To calculate the heat flow per square meter, we use the following formula: \(Q = \frac{\Delta T}{R_{SI}}\) Where \(Q\) is the heat flow (in W/m²), \(\Delta T\) is the temperature difference, and \(R_{SI}\) is the \(R\) factor in SI units. The temperature difference is given by the difference between the inside and outside temperatures: \(\Delta T = 23.0^{\circ}\mathrm{C} - (-22.0^{\circ}\mathrm{C}) = 45.0\) K Now, we can calculate the heat flow: \(Q = \frac{45.0}{7.044} = 6.389\) W/m² So, the heat flow per square meter through the wall is approximately \(6.39\) W/m².

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Key Concepts

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

R Factor Insulation
Understanding the R factor in insulation is crucial for maintaining energy efficiency in buildings. The R factor represents thermal resistance, which is the measure of how well a material resists heat flow. The higher the R value, the better the insulating capability of the material. Therefore, a wall with an R factor of 40, as mentioned in the textbook exercise, is considered to offer excellent insulation for harsh climates.

In simple terms, thermal resistance works as a barrier that slows down the transfer of heat from a warmer area to a cooler one. Insulation materials with higher R values will reduce the amount of heat escaping during winter and entering during summer, thus maintaining a comfortable temperature indoors and decreasing the energy required for heating and cooling systems. It's important for students to recognize that insulation is not just about thickness; the type of material and its density also affect the R value.
SI Units Conversion
In physics, consistency in units is crucial for accurate calculations and comparisons. The International System of Units (SI) is the standard set of units used globally. In our exercise, the R factor native units are in \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) h/BTU. However, to align with SI units, we convert them to m² K/W. The conversion process involves knowledge of basic conversion factors and their appropriate usage.

To help students contextualize the conversion, here are the steps broken down: first, translate the length from feet to meters, then convert Fahrenheit degrees to Kelvins, and finally, switch from BTUs (British Thermal Units) per hour to joules per second, which is watts. By understanding how each unit relates to the SI units, students can tackle a variety of problems where a conversion is necessary, not limited to thermal resistance.
Heat Flow Calculation
Heat flow calculation is a fundamental concept that helps us understand how heat energy transfers through materials. The formula used in the exercise, \(Q = \frac{\Delta T}{R_{SI}}\), illustrates the inverse relationship between heat flow and thermal resistance; as thermal resistance increases, the heat flow decreases. This is why materials with a high R value are preferred for insulation purposes.

The temperature difference (\(\Delta T\)) in the equation is the driving force for heat flow. It is the variance between the temperature inside and outside, in this case, represented in Celsius. When students calculate heat flow, the result expresses the amount of heat energy that passes through a specific area of material (per square meter) in a given time, hence the unit watts per square meter (W/m²). Understanding the dynamics of heat flow helps in designing more energy-efficient homes and can also be applied to various engineering problems regarding thermal management.

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Most popular questions from this chapter

A material has mass density \(\rho\), volume \(V\), and specific heat \(c\). Which of the following is a correct expression for the heat exchange that occurs when the material's temperature changes by \(\Delta T\) in degrees Celsius? a) \((\rho c / V) \Delta T\) b) \((\rho c V)(\Delta T+273.15)\) c) \((\rho c V) / \Delta T\) d) \(\rho c V \Delta T\)

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and waits for the system to reach equilibrium, when he will measure the temperature. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}, L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\), \(c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?

Enhanced geothermal systems (EGS) consist of two or more boreholes that extend several kilometers below ground level into the hot bedrock. Since drilling these holes can cost millions, one concern is that the heat provided by the bedrock cannot pay back the initial investment. How long can \(0.669 \mathrm{~km}^{3}\) of granite deliver an average of \(13.9 \mathrm{MW}\) of power, if its initial temperature is \(168.3^{\circ} \mathrm{C}\) and its final temperature is \(103.5^{\circ} \mathrm{C}\) ? [The density of granite is 2.75 times that of water, and its specific heat is \(0.790 \mathrm{~kJ} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right)\).

A girl has discovered a very old U.S. silver dollar and is holding it tightly in her little hands. Suppose that she put the silver dollar on the wooden (insulating) surface of a table, and then a friend came in from outside and placed on top of the silver dollar an equally old penny that she just found in the snow, where it had been left all night. Estimate the final equilibrium temperature of the system of the two coins in thermal contact.

In one of your rigorous workout sessions, you lost \(150 . \mathrm{g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming that the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg} .\) b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost ( 1 food calorie \(=4186 \mathrm{~J}\) ).

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