Question

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When used to boil water, the surface of the bottom exposed to the water is nominally at 110°C. If heat is transferred from the stove to the pan at a rate of 600 W, what is the temperature of the surface in contact with the stove for each of the two materials?

Answer 1

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$