∙A picture window has dimensions of 1.40 m ×2.50 m and is made of glass 5.20 mm thick. On a wint

step 1 Alright guys, let's look at the question. Question 1 was asking us to figure out the rate of hea

∙A picture window has dimensions of 1.40 m ×2.50 m and is...

$\bullet$ A picture window has dimensions of 1.40 $\mathrm{m} \times 2.50 \mathrm{m}$ and is made of glass 5.20 $\mathrm{mm}$ thick. On a winter day, the outside temperature is $-20.0^{\circ} \mathrm{C}$ , while the inside temperature is a comfortable $19.56^{\circ} \mathrm{C}$ (a) At what rate is heat being lost through the window by conduction? (b) At what rate would heat be lost through the window if you covered it with a $0.750-\mathrm{mm}$ -thick layer of paper (thermal conductivity 0.0500 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) )$ ?

Key Concepts

Composite Materials and Thermal Resistance

When dealing with multiple layers of materials, each layer adds its own thermal resistance to the overall system. The total heat transfer rate is determined by considering the series combination of these resistances. This concept is crucial when analyzing systems like a window covered with an additional layer, where both materials contribute to impeding heat flow.

Cross-sectional Area

The cross-sectional area through which heat is conducted directly affects the total rate of heat transfer. A larger area provides a greater path for the heat to flow, thereby increasing the rate at which energy is transferred through a material.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It varies between materials and plays a critical role in calculations involving heat transfer. High thermal conductivity indicates that a material is efficient at transferring heat, while low conductivity means it is more insulating.

Fourier's Law of Heat Conduction

Fourier's law is the fundamental principle that governs the conduction of heat through a material. It states that the rate of heat transfer is proportional to the negative of the temperature gradient and the cross-sectional area, with thermal conductivity serving as the proportionality constant. This law is essential for determining how quickly heat moves through different materials.

Temperature Gradient

The temperature gradient is the difference in temperature across a distance within a material. It acts as the driving force for heat transfer in conduction problems, with larger temperature differences resulting in higher rates of heat flow according to Fourier's law.

Transcript

00:01 Alright guys, let's look at the question.

00:02 Question 1 was asking us to figure out the rate of heat loss for the window.

00:11 So window h is equal to k a times th minus tcol.

00:17 H is the rate of heat loss and k is the thermoc conductivity for the window.

00:23 And a is the surface area and th is the higher temperature and tc is a lower temperature.

00:28 That is the thickness of the window.

00:32 Since the thickness of the window is given and the higher temperature is given, which is 19 .59 .56 degrees celsius.

00:40 And the lower temperature is given, which is negative 20 degrees celsius.

00:45 And surface area is given as well, which is 1 .4 o meter times 2 .5 ommeter.

00:51 And a thermal conductivity is 0 .8, watts per meter times calckel.

00:56 And that's plugging all the values.

00:58 Which you see here is 0 .8 watt per meter times californ, times 1 .40 meter times 2 .5m .m .m.

01:08 Times 19 .56 degrees celsius minus negative 20 .0 degrees celsius over 5 .2 times 10 to negative 3 meter.

01:18 That will give us.

01:21 The rate of heat loss is 2 .1 times 10 to the power 4 watts.

01:26 For question b, now there's a paper covering outside of the room, which is covering the outer side of the window, which is the side of the window that touches the outside environment.

01:42 Okay? so the red line here is the paper, and the black line here is the inside of the window, which is the inside of the room.

01:54 So there should be a middle temperature between the paper and window.

01:58 Okay? you see the temperature outside is negative 20 .0 degrees celsius, and temperature inside is 19 .56 degrees celsius.

02:07 So due to the thermal conduction, the heat loss by the paper is supposed to equal to the heat gain by the glass.

02:16 If we, right here, you see it's f prime equals h.

02:21 H prime is the rate of heat loss of paper, and the h here is the rate of heat gain by the window.

02:29 If we expand the equation, we give us k -prime and 10x8 times 8 times t .h -prime minus t -c -prime...

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