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Chapter 17: Problem 60

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路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路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掳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

(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0\(^\circ\)F to 45.0\(^\circ\)F in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was 44.0\(^\circ\)F on January 23, 1916. The next day the temperature plummeted to -56\(^\circ\)F. What was the temperature change in Celsius degrees?

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 kW/m\(^2\). The distance from the earth to the sun is \(1.50 \times 10{^1}{^1} m\), and the radius of the sun is \(6.96 \times 10{^8} m\). (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

Consider a poor lost soul walking at 5 km/h on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 W, and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k'A{_s}{_k}{_i}{_n}(T{_a}{_i}{_r} - T{_s}{_k}{_i}{_n})\), where \(k'\) is 54 J/h \(\cdot\) C\(^\circ\) \(\cdot\) m\(^2\), the exposed skin area \(A{_s}{_k}{_i}{_n}\) is 1.5 m\(^2\), the air temperature \(T{_a}{_i}{_r} \)is 47\(^\circ\)C, and the skin temperature \(T{_s}{_k}{_i}{_n}\) is 36\(^\circ\)C; (iii) the skin absorbs radiant energy from the sun at a rate of 1400 W/m\(^2\); (iv) the skin absorbs radiant energy from the environment, which has temperature 47\(^\circ\)C. (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e\) = 1 and that the skin temperature is initially 36\(^\circ\)C. Which mechanism is the most important? (b) At what rate (in L/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at 36\(^\circ\)C is \(2.42 \times 10{^6}\) J/kg.) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 m\(^2\). What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

BIO Treatment for a Stroke. One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at 0\(^\circ\)C to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached 32.0\(^\circ\)C. To treat a 70.0-kg patient, what is the minimum amount of ice (at 0掳C) you need in the bath so that its temperature remains at 0掳C? The specific heat of the human body is 3480 J/kg \(\cdot\) C\(^\circ\), and recall that normal body temperature is 37.0\(^\circ\)C.

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.2 cm thick on the inside wall surface. The wood has \(k\) = 0.080 W/m \(\cdot\) K, and the Styrofoam has \(k\) = 0.027 W /m \(\cdot\) K. The interior surface temperature is 19.0\(^\circ\)C, and the exterior surface temperature is -10.0\(^\circ\)C. (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall?
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