One end of a uniform brass rod 20 cm long and `10 cm^2` cross-sectional area is kept at `100^@C`. The other end is in perfect thermal contact with another rod of identical cross-section and length 10 cm. The free end of this rod is kept in melting ice and when the steady state has been reached, it is found that 360 g of ice melts per hour. Calculate the thermal conductivity of the rod, given that the thermal conductivity of brass is `0.25 cal//s cm^@C and L = 80 cal//g`.

Explanation

Let the thermal conductivity of iron be `K=cal//s-cm^@C`. The thermal resistance of brass rod
`R_1=(l_1)/(K_1A)=(15cm)/((0.25cal)/(s-cm^@C)xx20cm^2)=3 s-^(@)C//cal`
And that of iron rod
`R_(2)=(l_(2))/(K_(2)A)=(8 cm)/((K cal)/(s-cm^(@)C)xx20 cm^(2))=(2)/(5K)s-^(@)C//cal`
Since, the two rods are arranged in series, their effective thermal resistance is given by
`R=R_1+R_2=(3+(2)/(5K))s-^@C//cal`
Now, rate of heat flow through the rods`=(DeltaQ)/(Deltat)`
But `=((684g))/(3600s)(80+(cal)/(g))=15.2 cal//s`
Since `(DeltaQ)/(Deltat)=((T_1-T_2))/(R ) :. R=((T_1-T_2))/(DeltaQ)//(Deltat)`
`(3+(2)/(5K))s^@C//cal=((100-0)^@C)/(15.2 cal//s`
`K=111.8xx10^-3 cal//s-cm^(-@)C`