Question

Four $2cm\times 2cm\times 2cm$ cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at ${10}^{\circ }C$.
(a) Find the temperature of the drink when thermal equilibrium is attained in it.
(b) If the ice cubs do not do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the capacity. Density of ice $=900kg/m^3$, density of the drink $=1000kg/m^3$, specific heat capacity of the drink 4200 J/kg-K, latent heat of fusion of ice $3.4\times {10}^{5}J/kg.$

Answer 1

Given:-

Number of ice cube $=4$

Volume of each ice cube$=2\times 2\times 2=8c{m}^2$

Density of ice$=900kg/m^3$

Total number of ice,

$\begin{array}{c}{m}_1=4\times 8\times {10}^{-6}\times 900\\ =288\times {10}^{-4}kg\end{array}$

$\left(a\right)$Latent heat of fusion of ice,$L_i=3.4\times {10}^{5}J/kg$

Density of the drink$=1000kg/m^3$

Volume of the drink$=200ml$

mass of the drink$=(200\times {10}^{-6})\propto 1000kg$

Let first check the heat released when temperature of $200ml$ charges

from $10C$ to $0C$

$H_w=(200\times {10}^{-6}\times 1000\times 4200\times (10-0)=8400J$

Heat required to change four $8c{m}^3$ ice cube into water

$\left(H_i\right)=m_i{L}_i=(288\times {10}^{-4})\times (3.4\times {10}^5)=9792J$

$\therefore$ the heat required for melting the four cube of the ice

is greater than the heat released by water.$(H_i>H_w),$ some ice

will remain solid and there will be equilibrium between ice and water.

$\left(b\right)$ Equilibrium temperature of the cube and the drink $=0C$

Let M be the mass of the melted ice

$H_w=(200\times {10}^{-6}\times 1000\times 4200\times (10-0)$

$=8400J$

$\therefore M\times (3.4\times {10}^5)=8400J$

$\therefore M=0.0247kg=25g$