Question

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00∘C by ice–water mixture. The rod has a length of 55.0 cm and a cross-sectional area of 1.00 cm2. The heat conducted by the rod melts a mass of 5.00 g of ice in a time of 10.0 min. Find the thermal conductivity k of the metal?

Answer 1

Answer:

153.0815W/m.K

Explanation:

Heat transferred in phase is changed is expressed as:

$Q=\pm mL$ (m-mass, Q-heat, L-Latent heat of phase change)

Latent heat is the heat required to change the phase of 1kg of the material.

#The rate of heat flow(by conduction)  per unit time:

$H=\frac{\bigtriangleup Q}{\bigtriangleup t}=kA\frac{T_H-T_C}{L}\\\\L_f=334\times10^3J/kj\\\\H_{ice}=\frac{Q_{ice}}{t}=\frac{m_{ice}L_f}{t}$ #Heat flowing through melting ice.

$H_{ice}=\frac{5.0\times10^-^3kg)334\times10^3J/kg)}{10\times60s}\\\\=2.7833J/s$

To solve for k:

$H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}$$H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}$$H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}\\\\k=\frac{H}{A\frac{T_H-T_C}{L}}=\frac{HL}{A(T_H-T_C)}$

$=\frac{(2.7833J/s\times 0.55)}{1.00\times10^-^4m^2\times 100K}\\\\=153.0815W/m.K$

The thermal conductivity k of the metal is 153.0815W/m.K