3. A 2.5-m-high, 4-m-wide, and 20-cm-thick wall of a house has a thermal resistance of 0.025°C/W. The thermal conductivity of the wall is (a) 0.8 W/mK (b) 1.2 W/mK (c) 3.4 W/mK (d) 5.2 W/mK (e) 8.0 W/mK
Added by Patricia R.
Close
Step 1
5 m, 4 m, and 20 cm respectively. Show more…
Show all steps
Your feedback will help us improve your experience
Mayukh Banik and 63 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Heat is transferred steadily through a $0.2-\mathrm{m}$ thick $8 \mathrm{m} \times 4 \mathrm{m}$ wall at a rate of $2.4 \mathrm{kW}$. The inner and outer surface temperatures of the wall are measured to be $15^{\circ} \mathrm{C}$ and $5^{\circ} \mathrm{C} .$ The average thermal conductivity of the wall is $(a) 0.002 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}$ $(b)0.75 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}$ $(c) 1.0 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}$ $(d) 1.5 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}$ $(e) 3.0 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}$
The surface of a furnace wall is at a temperature of 1400°C. The outside wall temperature is 50°C. The furnace wall is made of 30 cm of refractory material having a thermal conductivity of 1.3 W/m-K. The outside wall is steel, 2 cm thick with a thermal conductivity of 45 W/m-K. Calculate the thickness in meters of brick to be installed between the refractory material and steel if its thermal conductivity is 0.35 W/m-K and the heat loss is not to exceed 900 W/m^2.
Penny R.
Five walls of a house have different surface areas, insulation materials, and insulation thicknesses. Rank them in order of the rate of heat flow through the wall, greatest to smallest. Assume the same indoor and outdoor temperatures for each wall. (a) area $=120 \mathrm{m}^{2} ; 10 \mathrm{cm}$ thickness of insulation with thermal conductivity $0.030 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$ (b) area $=120 \mathrm{m}^{2} ; 15 \mathrm{cm}$ thickness of insulation with thermal conductivity 0.045 W/(m-K) (c) area $=180 \mathrm{m}^{2} ; 10 \mathrm{cm}$ thickness of insulation with thermal conductivity $0.045 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$ (d) area $=120 \mathrm{m}^{2} ; 10 \mathrm{cm}$ thickness of insulation with thermal conductivity $0.045 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$ (e) area $=180 \mathrm{m}^{2} ; 15 \mathrm{cm}$ thickness of insulation with thermal conductivity $0.030 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD