Question

$50g$ ice at ${{0}^{o}}C$ in insulator vessel, $50g$ water of ${{100}^{o}}C$ is mixed in it, then final temperature the mixture is (neglect the heat loss)

& \text{A}\text{. }{{10}^{o}}C \\\ & \text{B}\text{. }{{0}^{o}}<<{{T}_{m}}<{{20}^{o}}C \\\ & \text{C}\text{. }{{20}^{o}}C \\\ & \text{D}\text{. Above }{{20}^{o}}C \\\ \end{aligned}$$

Answer 1

When ice is mixed with water, a body will lose its heat and whatever be heat has loosed, the heat will be gained by ice. So the heat is gained to convert 50g of ice at ${{0}^{o}}C$ to 50g of water at ${{0}^{o}}C$. In this case, ice is changing its state therefore use a latent heat formula and water is just increasing temperature to a final degree so use a specific heat formula. Use the energy balance equation. Latent heat formula is used when solid gains heat and then converted into liquid.

Formula used:
Latent heat is given by
${{H}_{L}}=mL$
Where,
${{H}_{L}}$= thermal energy
m= mass of substance
L= heat of fusion or vaporization
Formula of specific heat is given by,
${{H}_{L}}=mc\Delta T$
Where,
c= Specific heat capacity
$\Delta T$ = Change in temperature

Complete step by step solution:
In this question, it is given that 50 gram of ice at ${{0}^{o}}C$ is mixed with 50 gram of ${{100}^{o}}C$ in an insulator vessel. Since the vessel is insulator so there won’t be any heat loss.
Now, when ice is mixed with water, a body will lose its heat and whatever heat has loosed, the heat will be gained by ice.
So the heat is gained to convert 50g of ice at ${{0}^{o}}C$ to 50g of water at ${{0}^{o}}C$ is called latent heat. Which is given by,
${{H}_{L}}=mL$
This is the latent heat which is absorbed by ice to convert 50g of ice at ${{0}^{o}}C$ to 50g of water at ${{0}^{o}}C$.
And we know that body A is losing heat. So heat loss is given by,
${{H}_{L}}=mc\vartriangle T$
Let ${{T}^{0}}C$ be the final temperature of mixture. Latent heat of ice is $89cal/g$. Specific heat of water is $1cal/g/{}^{0}C$.
We know that heat lost by water to reach ${{T}^{0}}C$ is equal to sum of heat gained by ice to change itself into water and heat gained by melted ice to raise its temperature at ${{T}^{0}}C$.
Mathematically the energy balanced equation is given by,

& {{m}_{w}}{{c}_{w}}\vartriangle T=mL+{{m}_{i}}{{c}_{w}}\vartriangle T \\\ & 50\times 1(100-T)=50\times 80+\times 50\times 1\times (T-0) \\\ & 4000+50T=5000-50T \\\ & T={{10}^{0}}C \\\ \end{aligned}$$ So the final temperature of the mixture is $${{10}^{0}}C$$. Therefore the correct option is (A). **Additional information:** Latent heat of a substance is the quality of heat required to change the state of a unit mass of the substance without changing its temperature. Specific heat of a substance is defined as the quantity of heat required to raise the temperature of a unit mass of a substance through one degree. **Note:** Students should know the concept of the balance equation. If any substance is changing its state then use latent heat formula. But if the substance is maintaining its original state but increasing its temperature then use a specific heat formula. Latent heat of ice is $$89cal/g$$. Specific heat of water is $$1cal/g/{}^{0}C$$. Unit of specific heat is $$cal/g/{}^{0}C$$ and the unit of latent heat is $$89cal/g$$.