Solve for Cold Surface Temp. of Slab with Heat Flow Rate 240 W

AI Thread Summary
To find the cold surface temperature of a slab with a heat flow rate of 240 W, the relevant parameters include the slab's area (0.83 m²), thickness (1.4 cm), and thermal conductivity (0.6 W/(m·K)). The heat transfer equation H = -kA(ΔT/Δx) is applied, where H is the heat flow rate. The calculation for temperature difference (ΔT) results in an unexpectedly high value of 305°C, indicating a possible error in unit conversion or application of the formula. The correct thickness should be expressed in meters (0.014 m) to ensure accurate calculations. Properly addressing these factors will yield the correct cold surface temperature.
yankees26an
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Homework Statement


The absolute value of the heat flow rate through a slab of area 0.83 m2 is 240 W. The
thickness of the slab is 1.4 cm and its thermal conductivity is 0.6 W/(m·K). If the hot surface of the slab is at 40°C find the temperature of the cold surface.


Homework Equations



H = -kA(\DeltaT/\Deltax)

H = Q/t?

The Attempt at a Solution



A = 0.83 m^2; \Deltax = 1.4; k = 0.6 W/(m·K); Thot = 40°C;

Missing H?
 
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yankees26an said:

Homework Equations



H = -kA(\DeltaT/\Deltax)

H = Q/t?

this should be

\frac{dQ}{dt}=-kA \frac{d\theta}{dx}

or

ΔQ/Δt=-kA(Δθ/Δt)

you were given ΔQ/Δt as 240 W
 
ok so

\DeltaT = (\Deltax*H)/(kA)

\DeltaT = 1.2*240/0.6*0.83


\DeltaT = 578 K = 305 C ?? That's way above any answer options
 
yankees26an said:
\DeltaT = 1.2*240/0.6*0.83

this should be 0.014m
 
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