Answer to Question #197956 in Electric Circuits for Plea

Question

Water is boiling in an aluminum pan placed on an electrical element on a stovetop. The sauce pan has a bottom that is 0.800 cm thick and 14.0 cm in diameter. The boiling water is evaporating at the rate of 1.00 g/s. What is the temperature difference across (through) the bottom of the pan?

Answer 1

Gives

Thickness(d)=0.800Cm $=8\times10^{-3}m$

Diameter=14.0 cm

Radius=$7\times10^{-2}m$

$\frac{m}{t}=1.0\times10^{-3}kg/sec$

K_Al=220J/sec.m°c

L=2256$\times10^{3}J/kg$

$\frac{dQ}{dt}=\frac {kA({T_2-T_1)}}{d}$

Q=mL

$T_2-T_1=\frac{\frac{dQ}{dt}\times d}{kA}$

$T_2-T_1=\frac{L\frac{dm}{dt}\times d}{kA}\rightarrow(1)$

Put Value

$T_2-T_1=\frac{2256\times10^3\times\frac{10^{-3}}{1}\times 8\times10^{-3}}{220\times3.14\times(7\times10^{-2})^2}$

$T_2-T_1=5.33°C$