Question

When 2kg of ice $$-{20}^{\circ }C$$is mixed with $5Kg$ water ${20}^{\circ }C$ in an insulating vessel having negligible heat capacity. Calculate the final mass of water left in the container.
Given: Specific heats of water and ice are 1$$kcal{\left(k{g}^{\circ }C\right)}^{-1}$$ and $0.5$$$kcal{\left(k{g}^{\circ }C\right)}^{-1}$$latent heat of fusion of ice is 80$$kcal{\left(kg\right)}^{-1}$$
a. 5kg
b. 6kg
c. 7kg
d. 8kg

Answer 1

Hint: The temperature of Ice will start increasing and the temperature will start decreasing as soon as they are mixed.
The water will release energy and ice will consume that energy till both of them attain the same temperature.
When both of them attain the same temperature the remaining amount of heat will be consumed by ice to convert its state from solid to liquid.

Complete step by step answer:
Here it is given that $2Kg$ ice $$-{20}^{\circ }C$$is mixed with $5Kg$ water ${20}^{\circ }C$ in a container with no heat capacity this means that the container cannot store any amount of heat from the mixture
Now as we know the temperature of ice is lower and the temperature of the water is higher so heat will flow from water to ice i.e., the water will release energy and ice will take up that energy to change its state.
Now as we know at $0^{\circ }C$ ice and water both coexist.
Now we will calculate the amount of energy released by water after it decreases its temperature to $0^{\circ }C$
The amount of energy released is given as
$Q=mC\mathrm{\Delta }T$
Where
$m$is the mass of water
$C$ is the specific heat of the water
And $\mathrm{\Delta }T$ is the change in temperature
The given values are
$m=5Kg$
$C=1$$$kcal{\left(k{g}^{\circ }C\right)}^{-1}$$
$\mathrm{\Delta }T=20-0$
$\therefore Q=5\times 1\times 20J$
$\Rightarrow Q_{released}=100J$
So, the amount of heated released by water is $Q_{released}=100J$

Now we will calculate the amount of heat absorbed by ice to increase its temperature to $$0^{\circ }C$$
The amount of heat absorbed by the ice is
$Q=m_{ice}{C}_{ice}\mathrm{\Delta }T$
Where
$m_{ice}$ is mass of ice
$C_{ice}$ is the specific heat of ice
And $\mathrm{\Delta }T$ is the change in temperature of the ice

Here specific heat is used instead of latent heat because the state of ice is not changing only its temperature is increasing.
Here given values are
$m_{ice}=2Kg$
$C_{ice}=0.5$$$kcal{\left(k{g}^{\circ }C\right)}^{-1}$$
$\mathrm{\Delta }T=0-\left(-20\right)$
$\Rightarrow \mathrm{\Delta }T={20}^{\circ }C$
$\therefore Q=2\times 0.5\times 20J$
$\Rightarrow Q_{absorbed}=20J$

Now here we see that the energy released by the water is $Q_{released}=100J$
And the amount of energy absorbed by the ice is $Q_{absorbed}=20J$
So remaining heat will be consumed by ice to convert its state
Which is
$$Q_{remaining}=Q_{released}-Q_{absorbed}$$
$$\Rightarrow Q_{remaining}=100-20$$
$$\Rightarrow Q_{remaining}=80J$$
Now
We know that the heat absorbed by ice to convert its state is give
$Q_t=m_{ice}\left(L.H\right)$
Note here $Q_t=Q_{remaining}$
Where
$m_{ice}$ is the mass of ice converted into water
$\left(L.H\right)$ is the latent heat

So substituting values we get the mass of ice converted into water is
$\because Q_t=m_{ice}\left(L.H\right)$
$\therefore m={\displaystyle \frac{Q_1}{\left(L.H\right)}}$
Here the given values are
$$Q=80J$$
$$\left(L.H\right)=80$$$$kcal{\left(kg\right)}^{-1}$$
$$\therefore m={\displaystyle \frac{80}{80}}$$
$$\Rightarrow m=1kg$$
So the total mass of water in the mixture will be $5+1=6Kg$
The initial mass of ice converted into water $2-1=1kg$

Hence, the correct answer is option (B).

Note: Latent heat is the heat absorbed by the fluid to convert its state it does not have any effect on the temperature of the fluid
Specific heat is utilized by the fluid to increase its temperature
The flow of heat from higher to lower temperatures will continue till both the fluids attain the same temperature.