Question

The inner and outer surfaces of a \(0.5-\mathrm{cm}\) thick \(2-\mathrm{m} \times 2-\mathrm{m}\) window glass in winter are \(10^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the amount of heat loss through the glass over a period of \(5 \mathrm{~h}\). What would your answer be if the glass were \(1 \mathrm{~cm}\) thick?

Short answer

Question: Calculate the heat loss through a 2m x 2m glass window for each thickness of 0.5 cm and 1 cm over a period of 5 hours if the temperature difference between the inner and outer surfaces is 7°C and the thermal conductivity of the glass is 0.78 W/m·K. Answer: The heat loss through the 0.5 cm thick glass window is 34128 Ws and through the 1 cm thick glass window is 17064 Ws over a period of 5 hours.

Step-by-step solution

Identify the known variables

The thickness of the glass (L) is 0.5 cm which needs to be converted to meters, the surface area of the window is A = 2m * 2m, and the temperature difference between the inner and outer surfaces, ΔT = T1 - T2 = 10°C - 3°C = 7°C, which needs to be converted to Kelvin (ΔT = 7K). We are also given the thermal conductivity (k) of the glass, 0.78 W/m·K and the time of heat loss (t) = 5 hours.

Use the formula for heat conduction

The formula for heat conduction is Q = (k * A * ΔT * t) / L, where Q is the heat loss through the glass. First, we need to convert L to meters: L = 0.5 cm = 0.005 m, and the time to seconds: t = 5 h * 3600 s/h = 18000 s.

Calculate the heat loss for 0.5 cm thick glass

Now plug in the given values to the formula: Q = (0.78 * 4 * 7 * 18000) / 0.005. Calculate the result: Q = 34128 Ws.

Repeat steps for 1 cm thick glass

For 1 cm thick glass, L = 1 cm = 0.01 m. Use the same formula: Q' = (0.78 * 4 * 7 * 18000) / 0.01. Calculate the new result: Q' = 17064 Ws.

Interpret the results

The heat loss through the 0.5 cm thick glass over a period of 5 hours is 34128 Ws, and the heat loss through the 1cm thick glass is 17064 Ws. The thicker the glass, the less heat loss occurs due to its increased resistance to heat conduction.

Key concepts

Thermal Conductivity

Thermal conductivity is the measure of a material's ability to conduct heat. It determines how quickly heat can pass through a material when there is a temperature difference. Everyday materials like glass and metal have different thermal conductivities. For instance, the glass in our problem has a thermal conductivity of \(0.78 \, \mathrm{W/m \cdot K}\).

What this means is that for each square meter of surface area, a heat of \(0.78 \mathrm{~W}\) is conducted per degree difference in temperature, per meter thickness of the glass.

• High thermal conductivity means the material conducts heat well, like metal.
• Low thermal conductivity means the material is a good insulator, like wood or foam.

This property is crucial for determining how much heat loss will occur through the glass, which serves as an example of how thermal conductivity affects daily life, especially in energy management in buildings.

Heat Loss

Heat loss refers to the transfer of thermal energy from a warmer area to a cooler area. In our example, heat moves from the warm side of the glass at \(10^{\circ} \mathrm{C}\) to the cooler side at \(3^{\circ} \mathrm{C}\). Calculating heat loss helps in understanding how much energy is required to maintain desired temperature conditions.

The formula for calculating heat loss through a material is:\[ Q = \frac{k \cdot A \cdot \Delta T \cdot t}{L} \]where:

• \(Q\) is the heat loss,
• \(k\) is the thermal conductivity,
• \(A\) is the area through which heat is being conducted,
• \(\Delta T\) is the temperature difference,
• \(t\) is the time duration,
• \(L\) is the thickness of the material.

By using this formula, we can find that thinner glass allows for more heat to pass through, resulting in a higher rate of heat loss, as compared to thicker glass. This highlights the importance of understanding material properties to optimize energy use.

Temperature Difference

The temperature difference (\(\Delta T\)) between two points is a driving force for heat transfer. The greater the difference, the more heat will flow from the high-temperature region to the low-temperature region. In the given problem, the temperature difference across the glass is \(7^{\circ} \mathrm{C}\), which we handle in calculations as \(7 \mathrm{~K}\).

It's important to keep units consistent in your calculations. Whether in Celsius or Kelvin, the temperature difference remains the same, making it a straightforward conversion.

• A greater temperature difference increases the rate of heat transfer.
• In building design and energy management, managing temperature differences is key to efficient heating and cooling.

This concept is crucial in areas like insulation and climate control, where heat transfer efficiency needs managing to maintain comfortable and controlled environments.

Glass Thickness

Glass thickness influences the rate of heat transfer significantly. Thicker glass presents more material for heat to travel through, thus reducing the rate of heat loss. In the exercise, comparing a \(0.5\, \mathrm{cm}\) thick glass to a \(1\, \mathrm{cm}\) thick glass shows that doubling the thickness cuts the heat loss in half.

For the \(0.5\, \mathrm{cm}\) thick glass:\[ Q = \frac{0.78 \cdot 4 \cdot 7 \cdot 18000}{0.005} = 34128 \, ext{Ws}\]And for the \(1\, \mathrm{cm}\) thick glass:\[ Q' = \frac{0.78 \cdot 4 \cdot 7 \cdot 18000}{0.01} = 17064 \, ext{Ws}\]

• Thicker glass has a greater resistance to conduction.
• This resistance reduces energy costs by minimizing heat loss in winter or heat gain in summer.

Understanding how thickness affects heat transfer is crucial in areas such as glass window design, where energy efficiency can impact both environmental and economic factors.