Assuming all objects have the same emissivity (1.00), rank them based on the time rate at which they radiate heat (i.e. the thermal power) from the lowest power (first) to the highest power (last) given the surface area A and temperature T. T = -73 degrees Celsius A = 2.5 m^2 T = 27 degrees Celsius A = 0.8 m^2 T = 0 degrees Celsius A = 1.6 m^2 T = -23 degrees Celsius A = 0.8 m^2
Added by Tammy W.
Close
Step 1
15 = 200.15 K T2 = 27 + 273.15 = 300.15 K T3 = 0 + 273.15 = 273.15 K T4 = -23 + 273.15 = 250.15 K Show more…
Show all steps
Your feedback will help us improve your experience
Madhur L and 68 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
Six wood stoves have total surface areas $A$ and surface temperatures $T$ as given. Rank them in order of the power radiated, from greatest to least. Assume they all have the same emissivity. (a) $A=1.00 \mathrm{m}^{2}, T=227^{\circ} \mathrm{C}$ (b) $A=1.01 \mathrm{m}^{2}, T=227^{\circ} \mathrm{C}$ (c) $A=1.05 \mathrm{m}^{2}, T=227^{\circ} \mathrm{C}$ (d) $A=1.00 \mathrm{m}^{2}, T=232^{\circ} \mathrm{C}$ (e) $A=0.99 \mathrm{m}^{2}, T=232^{\circ} \mathrm{C}$ (f) $A=0.98 \mathrm{m}^{2}, T=232^{\circ} \mathrm{C}$
The emissivity of an object is 0.50. (a) Compared with a perfect blackbody at the same temperature, this object would radiate (1) more power, (2) the same amount of power, (3) less power. Why? (b) Calculate the ratio of the power radiated by the blackbody to that radiated by the object
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