Question

In the figure shown, $AB$ is a rod of length $30cm$ and area of cross-section $1.0c{m}^2$ and thermal conductivity $336SIunits$. The ends $A$ and $B$ are maintained at temperatures ${20}^{\circ }C$ and ${40}^{\circ }C$ respectively. A point $C$ of this rod is connected to a box $D$, containing ice at $0^{\circ }C$, through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is [Assume latent heat of fusion for ice, $L_f=80cal{g}^{-1})$

• A. $84mg{s}^{-1}$
• B. $84g{s}^{-1}$
• C. $20mg{s}^{-1}$
• D. $40mg{s}^{-1}$

Answer 1

The correct option is D $40mg{s}^{-1}$

Since $C$ and $D$ are connected through a highly conducting wire w/ no heat capacity, Temp and $C$ and $D$ will be same. If $Q$ is heat trans ferted to Ice
$Q=Q_{AC}+Q_{BC}$
$Q_{AC}=\frac{k{L}_{AC}}{A}(20-0)$
$Q_{BC}=\frac{k{L}_{BC}}{A}(40-0)$
$Q=\frac{k}{A}[L_A\times 20+L_{BC}\times 40]$
$=\frac{336}{1\times {10}^{-4}}[10\times {10}^{-2}\times 20+20\times {10}^{-2}\times 40]$
Melting rate $=\frac{Q}{L_i}\Rightarrow L_i=80\times 418J/g$
$=80\times 4180J/g$
$R=\frac{336}{{10}^{-4}}\left[\frac{1000\times {10}^{-2}}{80\times 4180}\right]$
$=0.1kg/s$
$Q=\frac{kA}{L}△T$
So $R=\frac{336}{80\times 4180}[\frac{20}{10}+\frac{40}{20}]\frac{{10}^{-4}}{{10}^{-2}}$
$=\frac{336\times 4\times {10}^{-2}}{80\times 4180}$
$=4\times {10}^{-5}kg/s$
$=40mg/s$