A liquid at 30°C is poured very slowly into a Calorimeter that is at temperature of 110°C. The boili

Q: Why doesn't the first 5g of liquid cool the calorimeter down below 80°C?

The problem states the liquid *completely* evaporates. Evaporation requires a massive amount of energy (latent heat) which is absorbed at the boiling point (80°C). Since it's poured 'very slowly,' thermal equilibrium is maintained. Once the calorimeter hits 80°C, it can no longer provide heat to boil any more liquid, so the process stops exactly at 80°C.

Q: What happens to the vapor after the first step? Do we include its heat capacity?

No, in open calorimeter problems where a liquid evaporates, the vapor escapes into the surrounding atmosphere, taking its thermal energy with it. It does not participate in the second phase of the experiment.

Q: Why do we equate the heat lost by the calorimeter in both steps?

It's a mathematical trick! In the first step, the calorimeter cools from 110°C to 80°C (a 30°C drop). In the second step, it cools from 80°C to 50°C (also a 30°C drop). Since the temperature drop is identical, the heat lost ($mx \Delta T$) is identical in both steps, allowing us to eliminate the unknown $mx$ term.

Q: Can we solve this if the temperature drops were different, say 40°C and 20°C?

Absolutely! You would just have two equations with $mx$ as a common variable. You could isolate $mx$ in one equation and substitute it into the other. The identical 30°C drop just makes the algebra slightly faster.

Q: Why is the latent heat so much larger than the specific heat?

Specific heat is the energy required to increase the kinetic energy of molecules (temperature). Latent heat is the energy required to completely break the intermolecular bonds to change phase from liquid to gas. Breaking bonds always requires significantly more energy than just making molecules vibrate faster.

Question

A liquid at $30^\circ	ext{C}$ is poured very slowly into a Calorimeter that is at temperature of $110^\circ	ext{C}$. The boiling temperature of the liquid is $80^\circ	ext{C}$. It is found that the first $5	ext{ gm}$ of the liquid completely evaporates. After pouring another $80	ext{ gm}$ of the liquid the equilibrium temperature is found to be $50^\circ	ext{C}$. The ratio of the Latent heat of the liquid to its specific heat will be ______$^\circ	ext{C}$. [Neglect the heat exchange with surrounding]

Sigma Percentile · Accepted Answer

The beauty of thermodynamics lies in its absolute adherence to the law of conservation of energy. In this thrilling problem, we are presented with a two-act play of heat exchange. We have a hot calorimeter and a cool liquid, but the liquid has a trick up its sleeve: it boils!

Let's dive into the mechanics of this thermal dance and uncover the hidden ratio of latent heat to specific heat.

### Analyzing the Setup

Imagine a sturdy calorimeter sitting on a table, radiating heat at a scorching $110^\circ	ext{C}$. We don't know its mass, let's call it $m$, and we don't know its specific heat capacity, let's call it $x$.

We are going to pour a liquid into it. The liquid starts at a cool $30^\circ	ext{C}$. We are given two crucial properties of this liquid: its specific heat capacity $s$, and its latent heat of vaporization $L$. The boiling point of this liquid is $80^\circ	ext{C}$.

The fundamental law governing our entire journey is the Principle of Calorimetry:

$$ 	ext{Heat Lost} = 	ext{Heat Gained} $$

### Phase 1: The Evaporation Act

In the first act, we slowly pour exactly $5	ext{ g}$ of the liquid into the hot calorimeter.

Because the calorimeter is at $110^\circ	ext{C}$ and the liquid boils at $80^\circ	ext{C}$, the liquid will first heat up from $30^\circ	ext{C}$ to $80^\circ	ext{C}$. But it doesn't stop there! The problem explicitly states that this $5	ext{ g}$ of liquid **completely evaporates**.

For the liquid to completely evaporate, the calorimeter must supply enough heat to not only raise the liquid's temperature but also to break its intermolecular bonds (the latent heat). Since the pouring is done "very slowly," the system reaches thermal equilibrium exactly at the boiling point. The calorimeter cools down to $80^\circ	ext{C}$, and the vapor escapes into the atmosphere.

Let's write the heat balance equation for this first phase:

$$ m \cdot x \cdot (110 - 80) = 5 \cdot s \cdot (80 - 30) + 5 \cdot L $$

Simplifying the temperature differences, we get:

$$ m \cdot x \cdot 30 = 5 \cdot s \cdot 50 + 5L $$

Which further simplifies to our first master equation:

$$ 30mx = 250s + 5L \quad \dots (i) $$

### Phase 2: Reaching Equilibrium

Now the curtain rises on the second act. The vapor from the first act has vanished. Our calorimeter is now sitting at $80^\circ	ext{C}$.

We pour in another $80	ext{ g}$ of the same liquid, which is again at $30^\circ	ext{C}$. This time, the liquid doesn't boil. The system simply exchanges heat until it reaches a final, peaceful equilibrium at $50^\circ	ext{C}$.

The calorimeter cools from $80^\circ	ext{C}$ to $50^\circ	ext{C}$, and the new batch of liquid warms from $30^\circ	ext{C}$ to $50^\circ	ext{C}$.

Let's write the heat balance equation for this second phase:

$$ m \cdot x \cdot (80 - 50) = 80 \cdot s \cdot (50 - 30) $$

Simplifying the temperature differences, we get:

$$ m \cdot x \cdot 30 = 80 \cdot s \cdot 20 $$

Which gives us our second master equation:

$$ 30mx = 1600s \quad \dots (ii) $$

### The Grand Finale: Equating the Heat

Look closely at equations $(i)$ and $(ii)$. Do you see the magic?

Both equations have the exact same term on the left side: $30mx$. This is because, in both phases, the calorimeter experienced exactly a $30^\circ	ext{C}$ drop in temperature (from $110^\circ	ext{C}$ to $80^\circ	ext{C}$, and then from $80^\circ	ext{C}$ to $50^\circ	ext{C}$).

This brilliant symmetry allows us to completely bypass the unknown mass and specific heat of the calorimeter! We can directly equate the right sides of both equations:

$$ 250s + 5L = 1600s $$

Now, it's just a matter of simple algebra. Let's group the terms with $s$ on one side:

$$ 5L = 1600s - 250s $$

$$ 5L = 1350s $$

We are looking for the ratio of the latent heat ($L$) to the specific heat ($s$). Dividing both sides by $5s$, we get:

$$ \frac{L}{s} = \frac{1350}{5} $$

$$ \frac{L}{s} = 270 $$

And there we have it! Through careful accounting of thermal energy and a beautiful mathematical cancellation, we found that the latent heat is exactly **270** times the specific heat.

Enter Numerical Value:

Visualized Solution

$V a r iab l es$

$Q_{lost} = Q_{gained}$

$m x Δ T = m_{1} s Δ T + m_{1} L$

$30 m x = 250 s + 5 L$

$m x Δ T = m_{2} s Δ T$

$30 m x = 1600 s$

$250 s + 5 L = 1600 s$

$5 L = 1350 s$

$\frac{L}{s} = 270$

$m_{v} L$

The Sigma Insight: Calorimetry

Analyzing the Setup

Phase 1

Phase 2

The Grand Finale

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qlost​=Qgained​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mxΔT=m1​sΔT+m1​L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 30mx=250s+5L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mxΔT=m2​sΔT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 30mx=1600s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 250s+5L=1600s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5L=1350s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sL​=270

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mv​L