Question

Heat is lost steadily through a \(0.5-\mathrm{cm}\) thick \(2 \mathrm{~m} \times 3 \mathrm{~m}\) window glass whose thermal conductivity is \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner and outer surface temperatures of the glass are measured to be \(12^{\circ} \mathrm{C}\) to \(9^{\circ} \mathrm{C}\). The rate of heat loss by conduction through the glass is (a) \(420 \mathrm{~W}\) (b) \(5040 \mathrm{~W}\) (c) \(17,600 \mathrm{~W}\) (d) \(1256 \mathrm{~W}\) (e) \(2520 \mathrm{~W}\)

Short answer

Answer: (e) 2520 W

Step-by-step solution

Identify the given information

We are given the following information for the window glass: - Thickness (t) = 0.5 cm = 0.005 m (converted to meters) - Area (A) = 2 m x 3 m = 6 m² - Thermal conductivity (k) = 0.7 W/m·K - Inner surface temperature (T_in) = 12°C - Outer surface temperature (T_out) = 9°C

Apply the heat conduction formula

The formula for heat conduction is: \(Q = \dfrac{k \times A \times (T_{in} - T_{out})}{t}\) where Q represents the rate of heat loss.

Calculate the rate of heat loss

Substitute the given values into the formula: \(Q = \dfrac{0.7 \cdot 6 \cdot (12-9)}{0.005}\) \(Q = \dfrac{4.2 \cdot 3}{0.005}\) \(Q = \dfrac{12.6}{0.005}\) \(Q = 2520 \mathrm{~W}\)

Match the calculation with the given options

The calculated rate of heat loss through the glass window is 2520 W, which matches with option (e). Therefore, the correct answer is: (e) \(2520 \mathrm{~W}\)

Key concepts

Thermal Conductivity

When examining how heat moves through materials, we encounter the term thermal conductivity, symbolized by 'k'. It quantifies a material's ability to conduct heat. The greater the thermal conductivity is, the more efficient a material is at transporting heat from its hotter part to its cooler part.

Using our example regarding a window glass with a thermal conductivity of 0.7 W/m·K, we can infer that glass is not an excellent conductor of heat, like metals are, but it's not a particularly bad one either. This property is crucial for materials used as insulators in buildings. Different materials have different thermal conductivities – for instance, a copper wire would possess a significantly higher thermal conductivity than glass, which means it would transfer heat much faster.

Understanding thermal conductivity helps in selecting appropriate materials for insulation or for any application where managing heat transfer is important.

Rate of Heat Loss

The rate of heat loss through a material is a measure of how quickly heat is transferred across it. By knowing the rate, we can estimate the energy efficiency of structures and the effectiveness of insulating materials. To calculate the rate, we generally require the material's thermal conductivity, the area through which heat is being transferred, the temperature difference, and the thickness of the material as demonstrated in the given window glass example.

The formula used to calculate the heat loss rate is:
\[Q = \dfrac{k \times A \times (T_{\text{in}} - T_{\text{out}})}{t}\]
where 'Q' is the rate of heat loss in watts (W), 'k' is the thermal conductivity, 'A' is the area, '\(T_{\text{in}}\)' and '\(T_{\text{out}}\)' are the internal and external temperatures, and 't' is the material's thickness.

Calculating the rate of heat loss is key for decisions concerning heating and cooling needs, and also for designing buildings to maximize energy conservation.

Heat Transfer Fundamentals

Heat transfer is a phenomenon that occurs when there is a temperature difference within a medium or between different media. There are three modes of heat transfer: conduction, convection, and radiation.

Conduction

This is the process we're focusing on with the window glass problem. It occurs when heat is transferred through a solid material as a result of a temperature gradient. The thermal energy moves from the warmer side to the cooler side through the vibration of molecules and the movement of electrons.

Convection

Involves the transfer of heat by the movement of fluids (gases or liquids). Hot fluids rise and cold fluids sink, which creates a cycle that transfers heat.

Radiation

The transfer of heat in the form of electromagnetic waves, which requires no medium and can even occur in a vacuum, such as heat from the sun reaching Earth.

Understanding the fundamentals of heat transfer allows us to control and manage heat for various applications, such as heating systems, thermal insulation, and even electronic devices that require efficient heat dissipation.