Question

A cube of side $10cm$ is filled with ice of density $0.9gm/cc$. The thickness of the walls of the cube is $1$ mm and thermal conductivity of the material of the cube is $0.01cal/cm{-}^{o}C$. If the cube is placed in steam bath maintained at a temperature of $100$${}^o$C , the time in which ice completely melts, is (latent heat of fusion of ice = $80$ cal/gm.)

• A. 6 sec
• B. 12 sec
• C. 24 sec
• D. 48 sec

Answer 1

Answer: (B)

12 sec

Given data:

Length of the cube $=10cm$

Density of ice$=0.9g/cc$

Thickness of the walls of the cube $=1mm$

Thermal conductivity of the material of cube $\left(K\right)=0.01cal/cm{-}^{o}C$

latent heat of fusion of ice $=80cal/gm.$

$1Kcal=4184Joules$

Now,

The mass of the ice in the cube is $m=\rho \times V=0.9\times {10}^3=900gm$

So total heat absorbed is $m\times 80cal/g=72\times {10}^{3}cal$

As there are 6 walls , rate of heat conducted would be

$Q^{.}=\frac{Q}{t}=\frac{K\times Area\times \Delta T}{L}$

$\Rightarrow$ $Q^{.}=(6\times 0.01\times 100\times 100)/0.1=6000cal/s$

So, time taken to melt the ice is $\frac{Q}{Q/t}=(72\times {10}^3)/6000=12sec$