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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

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

Step by step solution

01

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 \( k = 0.040 \, \text{W/m} \cdot \text{K} \).
02

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.
03

Substitute the Values into the Formula

Plug in the given values into the equation: \[\frac{Q}{t} = \frac{0.040 \, \text{W/m} \cdot \text{K} \times 1.40 \, \text{m}^2 \times (175 - 35) \, \text{K}}{0.04 \, \text{m}}.\]
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Rods of copper, brass, and steel-each with crosssectional area of 2.00 cm\(^2\)-are welded together to form a Y-shaped figure. The free end of the copper rod is maintained at 100.0\(^\circ\)C, and the free ends of the brass and steel rods at 0.0\(^\circ\)C. Assume that there is no heat loss from the surfaces of the rods. The lengths of the rods are: copper, 13.0 cm; brass, 18.0 cm; steel, 24.0 cm. What is (a) the temperature of the junction point; (b) the heat current in each of the three rods?

Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to 34.0\(^\circ\)C overnight and rise to 40.0\(^\circ\)C during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400-kg camel would have to drink if it attempted to keep its body temperature at a constant 34.0\(^\circ\)C by evaporation of sweat during the day (12 hours) instead of letting it rise to 40.0\(^\circ\)C. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 J/kg \(\cdot\) K. The heat of vaporization of water at 34\(^\circ\)C is \(2.42 \times 10{^6} J/kg\).)

A 25,000-kg subway train initially traveling at 15.5 m/s slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 m long by 20.0 m wide by 12.0 m high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 kg/m\(^3\) and its specific heat to be 1020 J /kg \(\cdot\) K.

A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other (Fig. E17.62). The rod consists of a 1.00-m section of copper (one end in boiling water) joined end to end to a length \(L_2\) of steel (one end in the ice-water mixture). Both sections of the rod have crosssectional areas of 4.00 cm\(^2\). The temperature of the copper- steel junction is 65.0\(^\circ\)C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length \(L_2\) of the steel section?

A picture window has dimensions of 1.40 m \(\times\) 2.50 mand is made of glass 5.20 mm thick. On a winter day, the temperature of the outside surface of the glass is -20.0\(^\circ\)C, while the temperature of the inside surface is a comfortable 19.5\(^\circ\)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-mm-thick layer of paper (thermal conductivity 0.0500 W/m \(\cdot\) K)?

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