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

[yankees26an]

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($$\Delta$$T/$$\Delta$$x)

H = Q/t?

The Attempt at a Solution

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

Missing H?

 

[rock.freak667]

Homework Equations

H = -kA($$\Delta$$T/$$\Delta$$x)

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

 

[yankees26an]

ok so

$$\Delta$$T = ($$\Delta$$x*H)/(kA)

$$\Delta$$T = 1.2*240/0.6*0.83


$$\Delta$$T = 578 K = 305 C ?? That's way above any answer options

 

[rock.freak667]

$$\Delta$$T = 1.2*240/0.6*0.83

this should be 0.014m