One end of an insulated metal rod is maintained at 100 ∘c, while the other end is maintained at 0∘c by an ice-water mixture. the rod is 60.3 cm long and has a cross-sectional area of 1.21 cm2. the heat conducted by the rod melts 8.55 g of ice in 10.0 min.

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

k= 236.29 W/mK

radiation warms things up temp rises convection currents air circ windy weather

pe = weightxheight = 2000x1000=2,000,000joules