00:01
Have ambient temperature t not equals to 300 kelvin and power generated p equals to 11 kilowatt and area a equals to 0 .50 meter square and shiny engine has emissivity e equals to 0 .050 whereas the engine that is painted block has emissivity e equals to 0 .95 determine the equilibrium temperature for the black engine and the shiny engine okay we have to calculate the temperatures okay for the black engine and for the shiny engine.
00:32
So when the body is placed in the some environment, then from the stephen boltzmann law, we can write that power p it is given by sigma e, and t to the power 4 minus t0 to the power 4.
00:49
So from here after rearranging we get temperature t, it will be equals to p by sigma a plus t0 to the power 4 and whole power 1 by.
01:00
4.
01:01
So now, calculating for the shiny engine, so temperature for the shiny engine, it will be equals to power p, which is 11 kilowatts, 11 into 10 to the power 3 divided by sigma, which is 5 .67 into 10 to the power minus 8 watt per meter square, kelvin power 4, and epsilon for the shiny engine, it is equals to 0 .050, and area it is 0 .50 plus 300 to the power 4, and whole power 1 by 4.
01:31
So from here after solving, we get temperature of the shiny engine comes out to be 11669 .5 kelvin, and this will be equals to this will be equals to 1 4 -0 degree centigrade.
01:48
So this is the answer for the temperature of the shiny engine.
01:52
Now, substituting values in this equation for the black engine, so we will get that temperature...