Question

The 5-mm-thick bottom of a 200-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:

Explanation:

Conduction of heat per second Q  through a metal sheet having thermal conductivity K , thickness d , area A and temperature difference between cold and hot surface ( T₂ -T₁ ) is given by the following equation

Q = $\frac{KA(T_2-T_1)}{d}$

Area A = π R² = 3.14 X 10000 X 10⁻⁶

= 3.14 X 10⁻2 m²

For aluminium plate :-

T₂ - T₁ = $\frac{Q\times d}{KA}$

= $\frac{600\times5\times10^{-3}}{240\times3.14\times10^{-2}}$

T₂ - T₁ = .4

T₂ = 110.4 ° C

For copper plate :

$[tex]\frac{600\times5\times10^{-3}}{390\times3.14\times10^{-2}}$[/tex]

T₂ - T₁ = .25

T₂ = 110.25 ° C