The emissivity of iron is approximately 0.25. If an iron cube (with a side length of 0.5 m) is heated to 200• C and is placed into a –10.C chamber then answer the following a) What is the net rate of heat loss due to radiation from the iron cube? b) If the cube is placed onto a 0.1 m thick aluminum sheet, and the ground of the chamber is assumed to be held at –10•C, how long will it take in seconds for the cube to lose 1000 J of heat via conduction to the ground? (Note: the thermal conductivity of aluminum is 240 W/(m K)).

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8th Edition
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Chapter1: Basic Modes Of Heat Transfer
Section: Chapter Questions
Problem 1.30P
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The emissivity of iron is approximately 0.25. If an iron cube (with a side length of 0.5 m) is
heated to 200• C and is placed into a -10•C chamber then answer the following
a) What is the net rate of heat loss due to radiation from the iron cube?
b) If the cube is placed onto a 0.1 m thick aluminum sheet, and the ground of the chamber
is assumed to be held at –10•C, how long will it take in seconds for the cube to lose 1000 J of heat via
conduction to the ground? (Note: the thermal conductivity of aluminum is 240 W/(m K)).
Transcribed Image Text:The emissivity of iron is approximately 0.25. If an iron cube (with a side length of 0.5 m) is heated to 200• C and is placed into a -10•C chamber then answer the following a) What is the net rate of heat loss due to radiation from the iron cube? b) If the cube is placed onto a 0.1 m thick aluminum sheet, and the ground of the chamber is assumed to be held at –10•C, how long will it take in seconds for the cube to lose 1000 J of heat via conduction to the ground? (Note: the thermal conductivity of aluminum is 240 W/(m K)).
c (J/(kg K)) | C (J/(mol K)) Lf (J/kg) | L. (J/kg) density (kg/m³)
19,300
8,960
2,700
7,500
1,000
790
Material
a (°C-1)
Gold
129
25
Copper
1.65×10-5
2.3x10-5
1.75×10-5
Aluminum
Stainless Steel
3.33x105
1.09×105
22.6x105
8.79×105
Water
4190
75
Ethyl Alcohol
2400
110
Ice
2090
37.6
917
1.01x105 Pa
1 atm =
NA = 6.022×1023 particles/mol
R = 8.3145 J/(mol K)
kB = 1.38×10-23 J/K
o = 5.67 x 10-8W/(m²K4)
Transcribed Image Text:c (J/(kg K)) | C (J/(mol K)) Lf (J/kg) | L. (J/kg) density (kg/m³) 19,300 8,960 2,700 7,500 1,000 790 Material a (°C-1) Gold 129 25 Copper 1.65×10-5 2.3x10-5 1.75×10-5 Aluminum Stainless Steel 3.33x105 1.09×105 22.6x105 8.79×105 Water 4190 75 Ethyl Alcohol 2400 110 Ice 2090 37.6 917 1.01x105 Pa 1 atm = NA = 6.022×1023 particles/mol R = 8.3145 J/(mol K) kB = 1.38×10-23 J/K o = 5.67 x 10-8W/(m²K4)
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