A wall is made of a substance whose thermal conductivity is .93 J / (m sec Celsius). What is the outside temperature of the wall if its thickness is 30 cm and its cross-sectional area 73 m^2, and if thermal energy flows through the wall at a rate of 30 watts when the inside temperature is 28 Celsius?
After attempting to work through this i think the change in temp is .1326 degrees but im not sure, also not sure if that would mean the outside temp is .1326 lower or higher than inside. please help
To find the outside temperature of the wall, you can use the formula for thermal conduction:
Rate of heat transfer (Q) = (Thermal conductivity × Area × Change in temperature) / Thickness
Given:
Thermal conductivity (k) = 0.93 J / (m sec Celsius)
Area (A) = 73 m^2
Thickness (L) = 30 cm = 0.3 m
Rate of heat transfer (Q) = 30 watts
Inside temperature (T1) = 28°C
First, let's convert the thickness from centimeters to meters:
Thickness (L) = 0.3 m
Now, rearrange the formula to solve for the change in temperature (ΔT):
ΔT = (Q × L) / (k × A)
Substituting the given values into the formula:
ΔT = (30 × 0.3) / (0.93 × 73)
ΔT ≈ 0.0128°C
This means that the change in temperature across the wall is approximately 0.0128°C. Since the inside temperature is 28°C, to find the outside temperature (T2), subtract the change in temperature from the inside temperature:
T2 = T1 - ΔT
T2 = 28°C - 0.0128°C
T2 ≈ 27.9872°C
Therefore, the outside temperature of the wall is approximately 27.9872°C.