Respuesta :
Answer:
True
The object with the temperature at 1200K emits twice as much radiation as the object at 1000K.
Explanation:
The Stefan-Boltzmann law establishes how much an object radiates as a consequence of its temperature.
[tex]R_{p} = \sigma \cdot T^{4}[/tex] (1)
Where [tex]R_{p}[/tex] is the radiant power, [tex]\sigma[/tex] is the Stefan-Boltzmann constant and T the temperature.
Case for the object with [tex]T = 1000K[/tex]:
[tex]R_{p} = (5.67x10^{-8} W/m^{2} K^{4})(1000K)^{4}[/tex]
[tex]R_{p} = 56700 W/m^{2}[/tex]
Case for the object with [tex]T = 1200K[/tex]:
[tex]R_{p} = (5.67x10^{-8} W/m^{2} K^{4})(1200K)^{4}[/tex]
[tex]R_{p} = 117573 W/m^{2}[/tex]
[tex]R_{p}diff = \frac{117573 W/m^{2}}{56700 W/m^{2}}[/tex]
[tex]R_{p}diff = 2.07[/tex]
Hence, the object with the temperature at 1200K emits twice as much radiation as the object at 1000K.
Answer:
True
P2/P1 ~= 2
Therefore, The object at 1200 K emits roughly twice as much radiation as the object at 1000 K
Explanation:
According to Stefan-Boltzmann law, which states that the amount of thermal energy radiated by an object per unit area per second is directly proportional to the fourth power of its absolute temperature. It can be stated mathematically as;
P/A = c.T^4 .......1
Where;
P = radiated power
A = surface area
c = stefan's constant
T = absolute temperature.
From 1;
P = cAT^4
For the two cases
P1 at T1 = 1000K
P2 at T2 = 1200K
Since the object is identical
A1 = A2 = A
P2/P1 = cA(T2)^4 ÷ cA(T1)^4
P2/P1 = (T2)^4/(T1)^4
P2/P1 = (1200)^4/(1000)^4
P2/P1 = 2.073
P2/P1 ~= 2
Therefore, The object at 1200 K emits roughly twice as much radiation as the object at 1000 K
