Respuesta :
Answer:
[tex]T_b=107.3784\ ^{\circ}C[/tex]
Explanation:
Given:
- thickness of the base of the kettle, [tex]dx=0.52\ cm=5.2\times 10^{-3}\ m[/tex]
- radius of the base of the kettle, [tex]r=0.12\ m[/tex]
- temperature of the top surface of the kettle base, [tex]T_t=100^{\circ}C[/tex]
- rate of heat transfer through the kettle to boil water, [tex]\dot Q=0.409\ kg.min^{-1}[/tex]
- We have the latent heat vaporization of water, [tex]L=2260\times 10^3\ J.kg^{-1}[/tex]
- and thermal conductivity of aluminium, [tex]k=240\ W.m^{-1}.K^{-1}[/tex]
So, the heat rate:
[tex]\dot Q=\frac{0.409\times 2260000}{60}[/tex]
[tex]\dot Q=15405.67\ W[/tex]
From the Fourier's law of conduction we have:
[tex]\dot Q=k.A.\frac{dT}{dx}[/tex]
[tex]\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}[/tex]
where:
[tex]A=[/tex] area of the surface through which conduction occurs
[tex]T_b=[/tex] temperature of the bottom surface
[tex]15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}[/tex]
[tex]T_b=107.3784\ ^{\circ}C[/tex] is the temperature of the bottom of the base surface of the kettle.
