Question

A cubical vessel of side 10 cm is filled with ice at ${0^0}C$ and is immersed in a water bath as ${100^0}C$. If thickness of walls of vessel is 0.2 cm and conductivity is 0.02 CGS units, then time in which all the ice melts is (Density of ice=0.9 gm/cc)
A. 431.1 sec
B. 9 sec
C. 12 sec
D. 15 sec

Answer 1

Generally when temperature difference is maintained then heat transfers from higher temperature body to the lower temperature body. The amount of heat transferred depends upon the various factors like the temperature difference, thermal resistance of the material. when heat is given to ice it converts into water and ideally when the same amount of heat is taken from water we can convert it back to ice.

Formula used:
\eqalign{ & \dfrac{{dQ}}{{dt}} = \dfrac{{\Delta T}}{R} \cr & R = \dfrac{L}{{KA}} \cr & {Q_L} = mL \cr}

Complete step by step answer:
Flow of a quantity with time is known as a current. It can be fluid current or heat current or electric current.
In case of fluid current a pressure difference is maintained and that drives the flow of fluid and fluid always flows from high pressure region to the low pressure region and fluid current is governed by fluid resistance too
Whereas in electric current the electric charge flows with time and the voltage difference and electric resistance combined will govern the electric current. Charge flows from higher voltage to lower voltage naturally.
Similarly in thermal current it is governed by temperature difference and thermal resistance.
Thermal resistance of a thermal conductor of length ‘L’ and cross sectional area ‘A’ and thermal conductivity ‘K’ is given by $R = \dfrac{L}{{KA}}$
Now, coming to latent heat of ice, It is the amount of heat taken in converting ice to water or the amount of heat given out when water is converted into ice.
Value of latent heat of ice is $80\dfrac{{calories}}{{gram}}$
The rate of generation of heat due to hot water bath at 100 degree centigrade is
$\dfrac{{dQ}}{{dt}} = \dfrac{{\Delta T}}{R}$
\eqalign{ & \Rightarrow \dfrac{{dQ}}{{dt}} = \dfrac{{100 - 0}}{{\left( {\dfrac{{6 \times 0.2}}{{0.02 \times {{10}^2}}}} \right)}} \cr & \Rightarrow \dfrac{{dQ}}{{dt}} = 167cal/\sec \cr & \Rightarrow {Q_{given}} = 167t \cr}
The reason why we multiplied the resistance with 6 is because there are six faces and heat enters from 6 faces.
The heat required for the melting of ice is its latent heat and its value is
${Q_L} = mL$
\eqalign{ & \Rightarrow {Q_L} = \rho VL \cr & \Rightarrow {Q_L} = (0.9)(1000)(80) \cr & \Rightarrow {Q_L} = 72000calories \cr & \therefore {Q_{required}} = 72000calories \cr}
So time required for melting Is
\eqalign{ & {Q_{given}} = {Q_{required}} \cr & \Rightarrow 167t = 72000 \cr & \Rightarrow t = \dfrac{{72000}}{{167}} \cr & \therefore t = 431.1\sec \cr}

So, the correct answer is “Option A”.

Note:
In case of calorimetry problems one usually confuses between latent heat and specific heat. Latent heat is the amount of heat taken which in turn causes phase change where temperature remains constant. But in case of specific heat phase remains constant while temperature changes.