Three containers C1, C2 and C3 have water at different temperatures. The table below shows the final

Q: Why can we use volume instead of mass in the equations?

Since all containers have water, the density $\rho$ and specific heat $s$ are the same. Mass $m = \rho \times V$. The $\rho$ and $s$ terms cancel out from the heat balance equation $\sum m_i s \Delta T_i = 0$, leaving only volume $V$.

Q: Do we need to find the exact values of $T_1$, $T_2$, and $T_3$?

No! That's the beauty of this problem. By adding the three equations, you directly get the sum $(T_1 + T_2 + T_3)$, which is exactly what you need for the final mixture equation.

Q: What if the specific heat changes with temperature?

In standard JEE problems, unless specified, we assume specific heat and density remain constant over the given temperature range to simplify the calculations.

Q: Why is the equation written as $V_1(T_1 - T_f) + V_2(T_2 - T_f) = 0$?

This represents the net heat exchange. If $T_1 > T_f$, the term is positive (heat lost). If $T_2 < T_f$, the term is negative (heat gained). The sum of all heat exchanges in an isolated system must be zero.

Q: Could we have solved this using a different method?

You could solve the system of three linear equations to find $T_1 = 80^\circ\text{C}$, $T_2 = 50^\circ\text{C}$, and $T_3 = 20^\circ\text{C}$, and then substitute them into the final equation. However, adding the equations is much faster and less prone to calculation errors.

Question

Three containers $C_1$, $C_2$ and $C_3$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

\begin{array}{|c|c|c|c|}
\hline
C_1 & C_2 & C_3 & T \
\hline
1l & 2l & - & 60^\circ	ext{C} \
- & 1l & 2l & 30^\circ	ext{C} \
2l & - & 1l & 60^\circ	ext{C} \
1l & 1l & 1l & 	heta \
\hline
\end{array}

The value of $	heta$ (in $^\circ	ext{C}$ to the nearest integer) is ......

Sigma Percentile · Accepted Answer

### The Elegance of Calorimetry: Mixing Water from Three Containers

Imagine you are in a laboratory with three distinct containers of water, labeled $C_1$, $C_2$, and $C_3$. Each container holds water at a different, unknown initial temperature. Let's call these temperatures $T_1$, $T_2$, and $T_3$. The problem presents us with a series of mixing experiments, and our goal is to predict the final temperature when equal volumes from all three containers are mixed together.

### The Principle of Calorimetry

The foundational tool we need here is the **Principle of Calorimetry**, which states that in an isolated system, the net heat exchange is always zero. In simpler terms, the heat lost by the hotter bodies perfectly equals the heat gained by the colder bodies.

Mathematically, this is expressed as:

$$\sum m_i s \Delta T_i = 0$$

Since we are mixing water with water, the density $ho$ and specific heat capacity $s$ remain constant. Because mass is the product of density and volume ($m = ho V$), the constants $ho$ and $s$ cancel out of our equation. This leaves us with a beautifully simplified relation based purely on volume:

$$\sum V_i (T_i - T_f) = 0$$

### Formulating the Equations

Let's translate the first three mixing processes from the table into mathematical equations.

**Process I:** We mix $1	ext{L}$ of $C_1$ and $2	ext{L}$ of $C_2$, resulting in a final temperature of $60^\circ	ext{C}$.

$$1 \cdot (T_1 - 60) + 2 \cdot (T_2 - 60) = 0$$
$$T_1 - 60 + 2T_2 - 120 = 0$$
$$T_1 + 2T_2 = 180 \quad \dots (1)$$

**Process II:** We mix $1	ext{L}$ of $C_2$ and $2	ext{L}$ of $C_3$, resulting in a final temperature of $30^\circ	ext{C}$.

$$1 \cdot (T_2 - 30) + 2 \cdot (T_3 - 30) = 0$$
$$T_2 - 30 + 2T_3 - 60 = 0$$
$$T_2 + 2T_3 = 90 \quad \dots (2)$$

**Process III:** We mix $2	ext{L}$ of $C_1$ and $1	ext{L}$ of $C_3$, resulting in a final temperature of $60^\circ	ext{C}$.

$$2 \cdot (T_1 - 60) + 1 \cdot (T_3 - 60) = 0$$
$$2T_1 - 120 + T_3 - 60 = 0$$
$$2T_1 + T_3 = 180 \quad \dots (3)$$

### The Mathematical Trick

Now, we have a system of three linear equations. A brute-force approach would be to solve for $T_1$, $T_2$, and $T_3$ individually. However, if we look closely at the structure of the equations, a much more elegant path reveals itself. Let's add all three equations together:

$$(T_1 + 2T_2) + (T_2 + 2T_3) + (2T_1 + T_3) = 180 + 90 + 180$$

Grouping the like terms, we get:

$$3T_1 + 3T_2 + 3T_3 = 450$$

Dividing the entire equation by 3, we find the sum of the three temperatures:

$$T_1 + T_2 + T_3 = 150 \quad \dots (4)$$

### The Final Act

In the final process, we mix $1	ext{L}$ from each of the three containers. Let the final equilibrium temperature be $	heta$. Setting up our calorimetry equation one last time:

$$1 \cdot (T_1 - 	heta) + 1 \cdot (T_2 - 	heta) + 1 \cdot (T_3 - 	heta) = 0$$

Expanding this gives:

$$T_1 + T_2 + T_3 - 3	heta = 0$$
$$3	heta = T_1 + T_2 + T_3$$

We already know from equation (4) that the sum of the temperatures is $150$. Substituting this value in:

$$3	heta = 150$$
$$	heta = 50^\circ	ext{C}$$

And there we have it! By leveraging the symmetry of the equations, we bypassed tedious algebra and arrived directly at the final answer. **The final temperature $	heta$ is $50^\circ	ext{C}$.**

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Principle of Calorimetry

Process I Equation

Process II Equation

Process III Equation

Adding the Equations

Process IV & Final Answer

The Sigma Insight: Calorimetry

The Elegance of Calorimetry

The Principle of Calorimetry

Formulating the Equations

The Mathematical Trick

The Final Act

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