An object is inside a room that has a constant temperature of 295 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object? Assume that the temperature of the object remains constant.
I thought it would be 295/3 or 295 X 3 but I a not sure.
radiated energy= constant*T^4
absorbed energy= constant*295^4
T^4=3*295^4
Temp= 295*sqrt (sqrt3)= ???
check my thinking, Hannah http://www.britannica.com/EBchecked/topic/564843/Stefan-Boltzmann-law
To find the temperature of the object, we can use the concept of thermal equilibrium. In thermal equilibrium, the power absorbed by an object is equal to the power emitted by the object.
Given that the object emits three times as much power as it absorbs from the room, we can say:
Power emitted by the object = 3 * Power absorbed by the object
Let's assume the power absorbed by the object is P. Therefore, the power emitted by the object is 3P.
We can use the Stefan-Boltzmann law to relate the power emitted by the object to its temperature:
Power emitted by the object = σ * A * T^4
Where:
- σ is the Stefan-Boltzmann constant (approximately 5.67 * 10^-8 W/(m^2 * K^4))
- A is the surface area of the object
- T is the temperature of the object in kelvins
Since the temperature of the object remains constant, we can assume that the surface area is constant as well. So, we can rewrite the equation as:
3P = σ * A * T^4
Now, let's compare this equation with the equation for the power absorbed by the object, which is given by:
Power absorbed by the object = σ * A * T_r^4
Where:
- T_r is the temperature of the room in kelvins
We can see that the area and the Stefan-Boltzmann constant are the same in both equations. So, we can equate the two:
3P = P
σ * A * T^4 = σ * A * T_r^4
Since σ * A is a constant, we can simplify the equation further:
T^4 = T_r^4
Taking the fourth root of both sides gives us:
T = T_r
Therefore, the temperature of the object is equal to the temperature of the room, which in this case is 295 K.