Question

A hollow metal cube, with side $0.5m$ and wall thickness $5 \times {10^{ - 3}}m$ is filled with ice. It is immersed in a water tank maintained at ${100^0}C$ . Calculate the amount of ice melted in $335\sec$ . (Conductivity of metal $= 0.5W{m^{ - 1}}{K^{ - 1}}$ , Latent heat of fusion of ice $= 335 \times {10^3}Jk{g^{ - 1}}$ )
A. $15kg$
B. $15g$
C. $1.5kg$
D. $1.5g$

Answer 1

Latent heat of fusion is the amount of energy required to change a solid substance into liquid without changing its temperature. As the hollow metal cube with ice is immersed in the water tank. The ice in the cube undergoes phase change. The energy required by the ice to convert into water and then energy required by the water to raise its temperature is given by the water tank.

Complete step by step answer:
The initial amount of the ice in the hollow metal cube is not given, let the mass of the ice melted be $m\,kg$ . We are given the value of latent heat of fusion as $L = 335 \times {10^3}Jk{g^{ - 1}}$ . Therefore, the amount of energy required for the fusion of ice to water $Q$ , will be:
$Q = mL$ --equation $1$
As the given hollow cube is of metal therefore, due to conduction there will be transfer of heat energy through this metal to the ice. The heat conducted across all the faces of the cube will be given as:
$H = \dfrac{{KA({\theta _2} - {\theta _1})t}}{x}$ -- equation $2$
$H$ is the amount of heat conducted across cube;
${\theta _2} - {\theta _1}$ is the temperature difference; ${\theta _2} - {\theta _1} = {100^0}C - {0^0}C = {100^0}C$
$t$ is the given time, $t = 335\sec$
$x$ is the thickness of wall; $x = 5 \times {10^{ - 3}}m$
$K$ is the conductivity of the metal; $K = 0.5W{m^{ - 1}}{K^{ - 1}}$
$A$ is the total surface area of the cube,
The cube has six surfaces, the surface area of each surface is ${\left( {0.5} \right)^2} = 0.25$
Therefore, the total surface area for six surfaces will be $A = 6 \times 0.25 = 1.5$
As there is no heat loss, the amount of energy required for the fusion of the ice must be equal to the amount of heat conducted across the cube. Therefore, from equation $1$ and equation $2$ , we have
$mL = \dfrac{{KA({\theta _2} - {\theta _1})t}}{x}$
$\Rightarrow m = \dfrac{{KA({\theta _2} - {\theta _1})t}}{{xL}}$
Substituting the values, we get
$m = \dfrac{{0.5 \times 1.5 \times 100 \times 335}}{{5 \times {{10}^{ - 3}}\left( {335 \times {{10}^3}} \right)}}$
$\Rightarrow m = 15\,kg$
The amount of ice melted will be $15\,kg$ .

So, the correct answer is “Option A”.

Note:
The total surface area of the cube through which the heat will be conducted will be six times the surface area of each surface.
All the units are in SI units.
Latent heat of fusion is the amount of heat energy required to convert ice into water without changing its temperature.