The temperature of equal masses of three different liquids x, y and z are 10°C, 20°C and 30°C, respe

Q: Why do we assume no heat is lost to the surroundings?

In standard calorimetry problems, unless a rate of heat loss or a container's heat capacity is explicitly mentioned, we assume an ideal insulated system where 100% of the heat lost by the hot object is absorbed by the cold object.

Q: Why did the mass 'm' just disappear from the equations?

The problem states 'equal masses of three different liquids'. Because the mass $m$ is the same for both the heat-gaining and heat-losing liquids, it appears as a common factor on both sides of the equation $ms_1\Delta T_1 = ms_2\Delta T_2$, allowing us to divide both sides by $m$ and cancel it out.

Q: Can we just average the temperatures to find the final mixture?

No! You can only take a simple average of temperatures if both the masses AND the specific heat capacities of the two liquids are identical. Here, the specific heats are different, meaning one liquid resists temperature change more than the other.

Q: What exactly does the ratio of specific heats tell us?

The ratio tells us how much more thermal energy one liquid needs to change its temperature compared to another. For example, $s_x/s_y = 2/3$ means liquid $x$ requires less heat to raise its temperature by $1^\circ\text{C}$ than liquid $y$ does.

Q: Why did we multiply the two ratios together?

We needed to find the relationship between liquid $x$ and liquid $z$ to solve the final equation. By multiplying $(s_x/s_y)$ by $(s_y/s_z)$, the $s_y$ terms cancel out algebraically, leaving us directly with $s_x/s_z$.

Question

The temperature of equal masses of three different liquids $x, y$ and $z$ are $10^\circ	ext{C}, 20^\circ	ext{C}$ and $30^\circ	ext{C}$, respectively. The temperature of mixture when $x$ is mixed with $y$ is $16^\circ	ext{C}$ and that when $y$ is mixed with $z$ is $26^\circ	ext{C}$. The temperature of mixture when $x$ and $z$ are mixed will be

Options:

(A) $28.32^\circ	ext{C}$

(B) $25.62^\circ	ext{C}$

(C) $23.84^\circ	ext{C}$

(D) $20.28^\circ	ext{C}$

Sigma Percentile · Accepted Answer

### The Setup: Three Liquids, Three Temperatures

Imagine you are in a laboratory with three identical beakers. Inside them are three different liquids: $x$, $y$, and $z$. The problem gives us a beautiful simplifying condition right at the start—all three liquids have the exact same mass, so $m_x = m_y = m_z = m$.

However, their thermal states are quite different. Liquid $x$ is sitting at a cool $10^\circ	ext{C}$, liquid $y$ is at $20^\circ	ext{C}$, and liquid $z$ is the warmest at $30^\circ	ext{C}$. Our goal is to figure out what happens when we mix the coldest liquid ($x$) with the hottest liquid ($z$). But to do that, we first need to understand how these liquids behave when they exchange heat.

### The First Mix: Finding the Heat Capacities

When we mix two liquids, the fundamental principle of calorimetry dictates that the heat lost by the hotter liquid must equal the heat gained by the colder liquid (assuming no heat is lost to the surroundings). The formula for heat exchange is $Q = ms\Delta T$, where $m$ is mass, $s$ is specific heat capacity, and $\Delta T$ is the change in temperature.

Let's analyze the first experiment: mixing liquid $x$ and liquid $y$. The final equilibrium temperature is given as $16^\circ	ext{C}$. 
Liquid $x$ warms up from $10^\circ	ext{C}$ to $16^\circ	ext{C}$, gaining heat. Liquid $y$ cools down from $20^\circ	ext{C}$ to $16^\circ	ext{C}$, losing heat. Setting up our calorimetry equation:

$$m s_x (16 - 10) = m s_y (20 - 16)$$

Because the masses are equal, the $m$ beautifully cancels out from both sides. We are left with:

$$s_x (6) = s_y (4)$$

Rearranging this gives us the ratio of their specific heats:

$$\frac{s_x}{s_y} = \frac{4}{6} = \frac{2}{3}$$

### The Second Mix: Completing the Puzzle

Now, let's look at the second experiment: mixing liquid $y$ and liquid $z$. The final temperature here settles at $26^\circ	ext{C}$. 
Liquid $y$ warms up from $20^\circ	ext{C}$ to $26^\circ	ext{C}$, while liquid $z$ cools down from $30^\circ	ext{C}$ to $26^\circ	ext{C}$. Applying the same principle:

$$m s_y (26 - 20) = m s_z (30 - 26)$$

Again, the mass $m$ cancels out, leaving us with:

$$s_y (6) = s_z (4)$$

This yields another specific heat ratio:

$$\frac{s_y}{s_z} = \frac{4}{6} = \frac{2}{3}$$

### The Final Mix: The Grand Finale

We are finally ready for the main event: mixing liquid $x$ and liquid $z$. To set up this equation, we need the relationship between $s_x$ and $s_z$. We can easily find this by multiplying the two ratios we just discovered:

$$\frac{s_x}{s_z} = \frac{s_x}{s_y} 	imes \frac{s_y}{s_z} = \frac{2}{3} 	imes \frac{2}{3} = \frac{4}{9}$$

Now, let's mix $x$ (at $10^\circ	ext{C}$) and $z$ (at $30^\circ	ext{C}$). Let the final unknown temperature be $T_3$. Liquid $x$ will gain heat, and liquid $z$ will lose heat:

$$m s_x (T_3 - 10) = m s_z (30 - T_3)$$

Cancel the mass $m$ and rearrange to isolate the specific heat ratio:

$$\frac{s_x}{s_z} = \frac{30 - T_3}{T_3 - 10}$$

Substitute our known ratio of $\frac{4}{9}$:

$$\frac{4}{9} = \frac{30 - T_3}{T_3 - 10}$$

Now, it's just a matter of simple algebra. Cross-multiply to solve for $T_3$:

$$4(T_3 - 10) = 9(30 - T_3)$$
$$4T_3 - 40 = 270 - 9T_3$$
$$13T_3 = 310$$
$$T_3 = \frac{310}{13} \approx 23.84^\circ	ext{C}$$

And there we have it! The final temperature of the mixture will be **$23.84^\circ	ext{C}$**. The elegance of this problem lies entirely in how the equal masses allow us to strip away the physical bulk of the liquids and focus purely on their intrinsic thermal properties.

Select Answer:

Visualized Solution

$m_{x} = m_{y} = m_{z} = m$

$Principle of Calorimetry$

$Mixing x and y$

$Ratio of Specific Heats (x, y)$

$Mixing y and z$

$Ratio of Specific Heats (y, z)$

$Ratio of Specific Heats (x, z)$

$Mixing x and z$

$Substituting the Ratio$

$Final Calculation$

$What if m_{x} \neq = m_{y} \neq = m_{z} ?$

The Sigma Insight: Calorimetry

The Setup

The First Mix

The Second Mix

The Final Mix

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Visualized Solution

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