Question

Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water boils away at a rate of 0.409 kg/min, what is the temperature of the lower surface of the bottom of the kettle? Assume that the top surface of the bottom of the kettle is at 100.0∘C.

Answer 1

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

• thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

• radius of the base of the kettle, $r=0.12\ m$

• temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

• rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

• We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

• and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

So, the heat rate:

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

From the Fourier's law of conduction we have:

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.