Question

$2Kg$ of ice at $-{15}^{\circ }C$ is mixed with $2.5Kg$ of water at ${25}^{\circ }C$ in an insulating container. If the specific heat capacities of ice and water are $0.5cal(g{}^{\circ }C{)}^{-1}$ and $1cal(g{}^{\circ }C{)}^{-1}$, 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, $m_w=2.5Kg=2500g$
Specific heat capacity of water, $c_w=1cal(g{}^{\circ }C{)}^{-1}$
Mass of ice, $m_i=2Kg=2000g$
Specific heat capacity of ice, $c_i=0.5cal(g{}^{\circ }C{)}^{-1}$
Energy released by water from temperature ${25}^{\circ }C$ to $0^{\circ }C$:
$Q_w=m_w{c}_w\mathrm{\Delta }T$
$⟹Q_w=2500\times 1\times 25$
$⟹Q_w=62500cal$
Energy gained by ice from temperature $-{15}^{\circ }C$ to $0^{\circ }C$:
$Q_i=m_i{c}_i\mathrm{\Delta }T$
$⟹Q_i=2000\times 0.5\times 15$
$⟹Q_i=15000cal$
Let m be the mass of ice which will change to water.
Latent heat of fusion of ice is $80cal(g{}^{\circ }C{)}^{-1}$.
Now remaining heat will be used for melting the ice.
$\text{Remaining Heat }=80m$
$62500–15000=80m$
We can evaluate the mass of ice which melts.
$m={\displaystyle \frac{62500–15000}{80}}=593.7g$
Total mass of water = $2500+593.7=3093.7g$
The amount of water present in the container (in kg nearest integer) = $3Kg$

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.