Question

$2 Kg$ of ice at $-15^{\circ} C$ is mixed with $2.5 Kg$ of water at $25^{\circ} C$ in an insulating container. If the specific heat capacities of ice and water are $0.5 cal (g\ ^\circ C)^{-1}$ and $1 cal (g\ ^\circ C)^{-1}$, find the amount of water present in the container? (in kg nearest integer)

Answer 1

Hint:Mass of water and ice are given along their specific heat capacities. When both water and ice will mix, then water attains less temperature than $25^{\circ} C$. Energy released by water is used by ice in two ways, firstly, ice will reach from $-15^{\circ} C$ to $0^{\circ} C$ and secondly, the remaining heat will be used to melt the ice at $0^{\circ} C$.

Complete step-by-step solution:
Mass of water, $m_{w} = 2.5 Kg = 2500g$
Specific heat capacity of water, $c_{w} = 1 cal (g\ ^\circ C)^{-1}$
Mass of ice, $ m_{i} = 2 Kg = 2000g$
Specific heat capacity of ice, $c_{i} = 0.5 cal (g\ ^\circ C)^{-1}$
Energy released by water from temperature $25^{\circ} C$ to $0^{\circ} C$:
$Q_{w}= m_{w} c_{w} \Delta T $
$\implies Q_{w}= 2500 \times 1 \times 25 $
$\implies Q_{w}= 62500 cal $
Energy gained by ice from temperature $-15^{\circ} C$ to $0^{\circ} C$:
$Q_{i}= m_{i} c_{i} \Delta T $
$\implies Q_{i}= 2000 \times 0.5 \times 15 $
$\implies Q_{i}= 15000 cal $
Let m be the mass of ice which will change to water.
Latent heat of fusion of ice is $80 cal (g\ ^\circ C)^{-1}$.
Now remaining heat will be used for melting the ice.
$\text{Remaining Heat } = 80\ m$
$62500 – 15000 = 80 m$
We can evaluate the mass of ice which melts.
$m = \dfrac{62500 – 15000 }{80} = 593.7 g$
Total mass of water = $2500 + 593.7 = 3093.7 g$
The amount of water present in the container (in kg nearest integer) = $3 Kg$

Note: Sensible heat is observed in a process as there is a change in the body's temperature. Latent heat is energy transported in a process without changing the body's temperature, for example, in a state change. Both sensible and latent heats are recognized in many processes of transfer of energy in nature.