Physics: 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.

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Water is being boiled in an open kettle that has

Ответ:

JGottem6489

18.01.2022

Physics

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

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:

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