Animated Solution for Physics - Properties of Solids and Liquids: When $M_1$ gram of ice at $-10^{\circ }\text{C}$ (specific heat = 0.5 \text{ cal g}^{-1} ^\circ\text{C}^{-1}) is added to $M_2$ gram of water at $50^{\circ }\text{C}$, finally no ice is left and the water is at $0^{\circ }\text{C}$. The value of latent heat of ice, in ${\text{cal g}}^{-1}$ is

The principle of calorimetry is one of the most elegant and intuitive concepts in thermodynamics. It simply states that in an isolated system, the heat lost by the hotter bodies must exactly equal the heat gained by the colder bodies. It is the universe's way of balancing the thermal checkbook!

Imagine you are standing in a lab with a beaker. Inside this beaker, you have $M_2$ grams of water sitting comfortably at $50^{\circ }\text{C}$. This is our "hot body."

Now, you drop in $M_1$ grams of ice that has been chilling at $-10^{\circ }\text{C}$. This is our "cold body." The problem tells us that eventually, all the ice melts, and the entire mixture settles at exactly $0^{\circ }\text{C}$.

Water Cooling Down

Let's look at the water first. It starts at $50^{\circ }\text{C}$ and cools down to $0^{\circ }\text{C}$. The heat it loses is sensible heat, meaning it causes a temperature change without a phase change.

The formula for sensible heat is:

For our water, the mass is $M_2$, the specific heat $s$ in CGS units is 1 \text{ cal g}^{-1} ^\circ\text{C}^{-1}, and the change in temperature $\Delta T$ is $(50-0)=50^{\circ }\text{C}$.

Substituting these values, the heat lost by the water is:

Ice Warming Up and Melting

Now, let's shift our focus to the ice. This is where many students make a critical mistake! Ice at $-10^{\circ }\text{C}$ cannot just instantly melt. It must first absorb enough heat to warm itself up to its melting point, which is $0^{\circ }\text{C}$. Only after reaching $0^{\circ }\text{C}$ can it begin to absorb latent heat to change its state from solid to liquid.

So, the heat gained by the ice happens in two distinct stages: 1. Warming the ice: From $-10^{\circ }\text{C}$ to $0^{\circ }\text{C}$. 2. Melting the ice: At $0^{\circ }\text{C}$.

The heat required to warm the ice is:

The heat required to melt the ice is:

Therefore, the total heat gained by the ice is the sum of these two:

Now, we bring it all together using the principle of calorimetry. The heat lost by the water must equal the heat gained by the ice.

Our goal is to find the latent heat of ice, $L$. Let's isolate $L$ algebraically. First, we can move the $5M_1$ term to the other side:

Finally, we divide the entire equation by $M_1$:

And there we have it! The latent heat of ice in this system is exactly $\frac{50M_2}{M_1}-5$. This perfectly matches option (a). Always remember to break down the heat transfer into individual, logical steps, especially when phase changes are involved!