Animated Solution for Physics - Properties of Solids and Liquids: When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C will be

Imagine you are a detective trying to solve a thermal mystery. You have two liquids, A and B, and you don't know their specific heat capacities. However, you do know how they behave when mixed. This is a classic application of the Principle of Calorimetry, which states that in an isolated system, the heat lost by the hotter object must exactly equal the heat gained by the colder object.

Mathematically, this is expressed as:

Here, $m$ represents mass, $S$ represents specific heat capacity, and $\Delta T$ represents the positive change in temperature.

In our first scenario, we are given a calibration experiment. We mix $100\text{ g}$ of liquid A at $100^{\circ }\text{C}$ with $50\text{ g}$ of liquid B at $75^{\circ }\text{C}$. The final equilibrium temperature settles at $90^{\circ }\text{C}$.

Let's plug these values into our master equation. Liquid A cools down from $100^{\circ }\text{C}$ to $90^{\circ }\text{C}$, so its temperature drop is $10^{\circ }\text{C}$. Liquid B warms up from $75^{\circ }\text{C}$ to $90^{\circ }\text{C}$, so its temperature rise is $15^{\circ }\text{C}$.

By dividing both sides by $250$, we uncover a beautiful, simplified ratio between the specific heats:

This ratio is our golden key. We don't need to know the exact values of $S_A$ and $S_B$; their relationship is enough to unlock the second part of the problem.

Now, we face the real question. We again take $100\text{ g}$ of liquid A at $100^{\circ }\text{C}$, but this time we mix it with $50\text{ g}$ of liquid B at a cooler $50^{\circ }\text{C}$. Let the unknown final temperature be $T$.

We set up the calorimetry equation just like before. Liquid A cools down to $T$, and liquid B warms up to $T$:

To make the algebra friendlier, let's divide both sides by $50$:

This equation looks intimidating because it has three unknowns: $S_A$, $S_B$, and $T$. But remember our golden key? We can substitute $S_A=\frac{3}{4}{S}_B$ into the equation:

Notice the magic? The specific heat $S_B$ appears on both sides as a multiplier. We can completely cancel it out!

Now, we just have a simple linear equation in $T$. Let's multiply both sides by $2$ to clear the fraction:

Finally, we group the $T$ terms on one side and the constants on the other:

The final equilibrium temperature of the mixture is exactly $80^{\circ }\text{C}$. By using the first mixture to find the ratio of specific heats, we elegantly bypassed the need for their absolute values.