# Video: Rate of Conductive Heat Transfer

A glass window has an area of 6.00 m² and a thickness of 0.500 cm. The inner surface of the window is at a temperature of 10.0°C and the outer surface of the window is at a temperature of −18.0°C. Determine the rate of heat loss through the window. Use a value of 0.840 w/m ⋅ °C for the thermal conductivity of glass.

07:21

### Video Transcript

A glass window has an area of 6.00 meters squared and a thickness of 0.500 centimeters. The inner surface of the window is at a temperature of 10.0 degrees Celsius, and the outer surface of the window is at a temperature of negative 18.0 degrees Celsius. Determine the rate of heat loss through the window. Use a value of 0.840 watts per meter degrees Celsius for the thermal conductivity of glass.

Okay, so in this question, what do we have? We have a glass window, and this glass window has an area of 6.00 meters squared and a thickness of 0.500 centimeters. So let’s draw a diagram of the glass window. Now here is the window and actually it’s glass so let’s make it a little bit see-through. Here we go! It’s more see-through now. Okay, so we know the thickness of the window, so that’s this distance here. And let’s call that distance 𝑑. We know that distance 𝑑 is 0.500 centimeters because that’s the thickness of the window. Secondly, we also know the area of the window, so that’s this area here, which we will call 𝐴. We know the value of 𝐴, which is the area, is 6.00 meters squared. So that’s our window.

Now let’s read the rest of the question. So we’re told that the inner surface of the window is at a temperature of 10.0 degrees Celsius and the outer surface of the window is at a temperature of negative 18.0 degrees Celsius. So let’s imagine we’re looking at the window from the outside. And so the temperature on the left is the inside temperature, which we’ll call 𝑇 sub 𝑖. And the temperature on the right of the glass, the outside temperature, we’ll call 𝑇 sub 𝑜. Now we know the values of 𝑇 sub 𝑖 and 𝑇 sub 𝑜. 𝑇 sub 𝑖 is 10.0 degrees Celsius, and 𝑇 sub 𝑜, the outside temperature, is negative 18.0 degrees Celsius. Now what we’re asked to do in the question is to work out the rate of heat loss through the window.

The final piece of information that we’re given is a value of 0.840 watts per meter degrees Celsius for the thermal conductivity of glass. So we’re saying that this glass window has a thermal conductivity, which will call 𝜅. And we know what 𝜅 is. 𝜅 has been given to us as 0.840 watts per meter degrees Celsius. So we’ve got all the information we can from the question, which means we can erase the question now. Great! Now we’ve got space to work out some stuff. So what stuff do we need to work out? Well we have been asked to find out the rate of heat loss through the window. So how are we going to go about doing that? Well we can recall an equation that links together all of the quantities that we’ve been given in the question. The equation is the following: the heat transferred 𝑄 divided by the time taken for heat transfer to occur 𝑡 is equal to the thermal conductivity of the material through which heat is transferred multiplied by the area the surface area of the material multiplied by the temperature difference across the material divided by the thickness of the material.

Now if we look at the values carefully that we’ve been given, we’ve got everything on the right-hand side. We know 𝜅; we know area; we know 𝑑; and we can work out what Δ𝑇 is because we know the temperature difference across the window in this case. What we don’t know are the two quantities on the left-hand side: the heat transferred or the time taken for the heat to transfer. However, the question is asking us to work out the rate of heat loss. In other words, what the question is asking us to work out is the whole of this left-hand side. We don’t want to work out individually what 𝑄 is or what 𝑡 is. What we want to work out is 𝑄 over 𝑡, the amount of heat transferred per unit time or heat loss divided by time, which in other words is the rate of heat loss, which means that all we need to do now is to sub in all the values on the right- hand side.

So let’s go back doing that. But before we do, we need to notice something. The thickness 𝑑 has been given to us in centimeters. However, we want to convert this to the standard unit of meters. So to do this, we need to recall that one centimeter is equal to 10 to the power of negative two meters. Therefore, the value of 𝑑 that we have is 0.500 centimeters. And that’s the same as 0.500 times 10 to the power of negative two meters because that’s what one centimeters is. And therefore, we can replace our value of 𝑑 with a value in meters. So we’re converting things to standard units. Well, are we going to convert 𝑇 sub 𝑖 and 𝑇 sub 𝑜 the temperatures into Kelvin? Well, no we’re not going to do that because what we’re interested in is not the temperatures themselves but the difference in temperature. In other words, what we’re studying here is Δ𝑇, and Δ𝑇 is the same as 𝑇 sub 𝑖 minus 𝑇 sub 𝑜.

Now when we take the difference in two temperatures in Celsius, then this value is the same as if we took the temperature difference between the same temperatures in Kelvin. In other words, if we converted 𝑇 sub 𝑖 into Kelvin and 𝑇 sub 𝑜 into Kelvin and then we found Δ𝑇 is equal to 𝑇 sub 𝑖 minus 𝑇 sub 𝑜 in Kelvin, we get the same answer. The other thing is our value of 𝜅, the thermal conductivity of glass, has been given to us in watts per meter degrees Celsius. So we can afford to keep the temperatures in degrees Celsius. Anyway, so let’s find out what 𝑄 over 𝑡 is. We find that the rate of heat loss through the window 𝑄 over 𝑡 is equal to 𝜅 multiplied by the area 𝐴 multiplied by the difference in temperature Δ𝑇, which we said earlier was 𝑇 sub 𝑖 minus 𝑇 sub 𝑜. And this whole thing is divided by the value of 𝑑. And what we can do is to consider very quickly the units.

So we’ve got this one power of m in the denominator which takes out one power in the numerator here. And the other one here in the denominator takes out the second power of m. Secondly, we’ve also got degree Celsius here in the denominator which cancels with the unit of temperature difference, which will also be degree Celsius once we’ve computed what 10.0 degrees Celsius minus negative 18.0 degrees Celsius is. And so the units that we’re left with is watts. And this makes sense. We’re calculating the rate of transfer of heat. And heat has unit of energy, and time obviously has the units of time. And we know that power is the rate of change of energy. In other words, heat has units of joules and time has units of seconds, and joules per second is the same as watts.

So dimensionally, we’re doing well. Now all that’s left is to compute the fraction on the right-hand side. And once we do this, we find that 𝑄 over 𝑡 is equal to 28224 watts. However, we can’t give a final answer like this. If we look at all the values that were given to us in the question, all of them have been given to us to three significant figures. Therefore, we need to give our answer to three significant figures as well. So significant figure number one, number two, number three, the next one is a two. And this is less than five. So this significant figure, the third one, will stay the same. It does not round up. And therefore to three significant figures, our final answer is 28200 watts. But we can give this in a slightly more user-friendly unit. Why should we write out a long five-digit number when we can give it in kilowatts? Now one kilowatt is the same as 1000 watts. So to find out this value in kilowatts, we need to divide it by 1000. Doing this gives us a value of 28.2 kilowatts. And so either of these answers are valid. The rate of heat loss through the window is either 28200 watts to three significant figures or 28.2 kilowatts to three significant figures.