Question

A metallic rod of cross-sectional area $20c{m}^2$, with the lateral surface insulated to prevent heat loss, has one end immersed in boiling water and the other in ice water mixture. The heat conducted through the rod melts the ice at the rate of $1$ gm for every $84$ sec. The thermal conductivity of the rod is $160W{m}^1{K}^1$. Latent heat of $ice=80cal/gm,1cal=4.2joule$. What is the length (in m) of the rod?

• A. $4m$
• B. $5m$
• C. $8m$
• D. $10m$

Answer 1

The correct option is C $8m$

Given - $A=20c{m}^2=20\times {10}^{-4}{m}^2,K=160W/m-K=160W/m{-}^{o}C,L=80cal./g,$

as one end of rod is immersed in boiling water and other is in ice water mixture , therefore temperature difference of the two ends will be ,

$\Delta \theta =100-0={100}^{o}C$ ,

let $l$ be the length of metallic rod .

The rate of heat flow in rod is given by ,

$Q/t=KA\Delta \theta /l$ ..................eq1

,

now , the rate at which heat is gained by the ice will be ,

$Q^{\prime }/t=80/84cal./s=80\times 4.2/84J/s$ ...................eq2 ,

as the heat is transferred to the ice by metallic rod , therefore rate of heat flow in rod would be equal to the heat gained by ice i.e. eq1 and eq2 will be equal , hence

$KA\Delta \theta /l=80\times 4.2/84$ ,

or $l=\frac{KA\Delta \theta \times 84}{80\times 4.2}=\frac{160\times 20\times {10}^{-4}\times 100\times 84}{80\times 4.2}=8m$