Finding an objects surrounding temperature

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To find the surrounding temperature (T0) of an object using the given parameters, the equation P=εσA(T^4-T0^4) is applicable. By rearranging the equation, T0 can be isolated as T0 = (T^4 - P/εσA)^1/4. Substituting the values A=1.5, T=303 K, σ=5.67x10^-8, ε=0.9, and P=155 W into the equation allows for the calculation of T0. The discussion emphasizes the importance of understanding the units, noting that power is expressed in Watts, equivalent to Joules per second. This approach provides a clear method for determining the surrounding temperature based on the object's radiative properties.
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Homework Statement



Hi, I have just started with physics and this is my first time posting on this forum, The question is very confusing. it asks:
What is an objects surrounding temperature (T0) if we know the:
A=1.5
T=303 K
σ=5.67x10^-8
ε=.9
P (net radiated power gained)= 155 W

See attachment, question 3(c)

Homework Equations



either;
P=εσA(T^4-T0^4) or εσAT^4

The Attempt at a Solution



I have not made an attempt, the answer of the first equation requires both surrounding temperature (T0) and the temperature of the object (T) but the answer would be the rate at which an object radiates energy; in joules, not watts (from what I understand) but the other equation doesn't have the T0, so you can't use algebra to solve it.
Please don't just type the answer, explain how you got it.

Thanks.
 

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You appear to have all the info needed to solve your equation for T0 using..

P=εσA(T^4-T0^4)

where P is in Watts. One Watt = one Joule per second.

P=εσAT^4 - εσAT0^4

εσAT0^4 = εσAT^4 - P

T0^4 = T^4 - P/εσA

T0 = (T^4 - P/εσA)^-4

Substitute..
A=1.5
T=303 K
σ=5.67x10^-8
ε=.9
P (net radiated power gained) = 155 W = 155 J/S
 
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