Answer:
k= 236.29 W/mK
Explanation:
Assuming we have to find the thermal conductivity K of the metal.
which is given by the formula
k = (∆E/(A•∆t)) / (∆T/∆x)
where ∆E= energy imparted for melting= mL
L= latent heat of melting = 333.5 J
∆E = 333.5×8.55 J
∆E = 2841.4 J
A = cross-sectional area = 1.21×10^{-4} m^2
∆t =time of heating= 10×60 = 600 s
∆T = 100° C
∆x=length of the rod= 0.603 m
now substituting the values we get
[tex]k= \frac{2845}{1.21\times10^{-4}\times600}\times\frac{0.603}{100}[/tex]
k= 236.29 W/mK
