Question

An electric kitchen range has a total wall area of 1.40 m\(^2\) and is insulated with a layer of fiberglass 4.00 cm thick. The inside surface of the fiberglass has a temperature of 175\(^\circ\)C, and its outside surface is at 35.0\(^\circ\)C. The fiberglass has a thermal conductivity of 0.040 W /m \(\cdot\) K. (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 m\(^2\)? (b) What electric-power input to the heating element is required to maintain this temperature?

Short answer

The heat current and electric-power input required are both 196 Watts.

Step-by-step solution

Identify the Given Variables

The problem provides several values we'll use: wall area \( A = 1.40 \, \text{m}^2 \), thickness of insulation \( d = 0.04 \, \text{m} \), temperatures \( T_1 = 175^{\circ}C \) and \( T_2 = 35^{\circ}C \), and thermal conductivity \( k = 0.040 \, \text{W/m} \cdot \text{K} \).

Set Up the Formula for Heat Current

The formula for heat current \( Q/t \) through a flat slab is given by: \[\frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{d}.\]This relates thermal conductivity, area, temperature difference, and thickness to find the heat current.

Substitute the Values into the Formula

Plug in the given values into the equation: \[\frac{Q}{t} = \frac{0.040 \, \text{W/m} \cdot \text{K} \times 1.40 \, \text{m}^2 \times (175 - 35) \, \text{K}}{0.04 \, \text{m}}.\]

Calculate the Heat Current

Perform the calculation: \[\frac{Q}{t} = \frac{0.040 \times 1.40 \times 140}{0.04} = 196 \, \text{W}.\]So, the heat current through the insulation is 196 Watts.

Determine the Electric Power Required

Since the heater needs to supply enough power to maintain a constant temperature, the electric power input must be equal to the heat current. Thus, the electric-power input required is 196 Watts.

Key concepts

Thermal Conductivity

Thermal conductivity is an essential property when it comes to understanding how materials conduct heat. It describes the ability of a substance to transfer heat through itself. In simple terms, it measures how quickly heat passes through a material.
In our exercise, the fiberglass insulation has a thermal conductivity of 0.040 W/m·K. This value indicates how effectively heat is transferred across the material. The lower the thermal conductivity, the slower the heat travels through the material.

• Units: The thermal conductivity is measured in Watts per meter per Kelvin (W/m·K).
• Key Factors: Factors affecting thermal conductivity include material composition, temperature, and structure.
• Insulator vs Conductor: Good thermal insulators have low thermal conductivity, while good conductors have high values.

Understanding thermal conductivity helps predict how well an insulator like fiberglass can prevent heat loss in applications such as a kitchen oven.

Heat Current

Heat current, also known as heat flow rate, quantifies the amount of heat transferred per unit time through a particular area. It is often denoted as \( \frac{Q}{t} \), where \( Q \) is the heat energy and \( t \) is the time.
In the exercise, we use the formula for heat current through a slab:\[\frac{Q}{t} = \frac{k \cdot A \cdot (T_1 - T_2)}{d},\]where:

• \( k \) is the thermal conductivity.
• \( A \) is the area through which heat is transferred.
• \( T_1 - T_2 \) is the temperature difference across the slab.
• \( d \) is the thickness of the material.

By calculating these values, we determine that the heat current through the insulation is 196 Watts, indicating how much heat is escaping per second. This value is crucial for figuring out how to maintain temperatures inside the oven.

Electric Power

Electric power refers to the rate at which electrical energy is consumed or transferred. In the context of our problem, it concerns the power supplied to maintain the desired temperature in the oven.
In the given task, we found that the heat current through the insulation was 196 Watts. To sustain the oven's internal temperature at 175°C, an electric power input of the same 196 Watts is necessary.

• Maintaining Balance: Electric power needs to match the heat lost to keep the internal temperature stable.
• Measured in Watts: Just like the heat current, electric power is also measured in Watts.

This principle highlights the relationship between electrical energy input and maintaining an energy balance in thermal systems, ensuring effective functioning and efficiency of appliances like electric ovens.