Two identical beakers A and B contain equal volumes of two different liquids at 60 degrees C each an

Q: Why do we assume the volume and area are the same?

The problem states the beakers are identical and contain equal volumes. This means the exposed surface area and total volume are identical for both.

Q: Why is the graph a curve and not a straight line?

Newton's Law of Cooling is a first-order differential equation. The rate of cooling depends on the temperature itself, which yields an exponential decay curve, not a linear drop.

Q: What is the physical meaning of the product of density and specific heat?

It represents the volumetric heat capacity, or thermal inertia per unit volume. A higher value means it takes more energy to change the temperature of a given volume.

Q: Does the initial temperature matter for the comparison?

No, because both start at the same temperature, the relative rates of cooling depend entirely on their thermal properties.

Q: What if beaker A had a higher emissivity?

A higher emissivity would increase the heat transfer coefficient, making beaker A lose heat and cool down even faster.

Question

Two identical beakers A and B contain equal volumes of two different liquids at $60^{\circ}	ext{C}$ each and left to cool down. Liquid in A has density of $8 	imes 10^2 	ext{ kg/m}^3$ and specific heat of $2000 	ext{ J kg}^{-1}	ext{K}^{-1}$ while liquid in B has density of $10^3 	ext{ kg m}^{-3}$ and specific heat of $4000 	ext{ J kg}^{-1}	ext{K}^{-1}$. Which of the following best describes their temperature versus time graph schematically? (Assume the emissivity of both the beakers to be the same)

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Beakers

Imagine you are in a laboratory. On the bench in front of you sit two identical beakers, labeled A and B. Both are filled with the exact same volume of liquid, and both have been heated to a precise $60^{\circ}	ext{C}$.

You step back and watch them cool. But here is the catch: the liquids inside are fundamentally different. Liquid A is slightly lighter, with a density of $8 	imes 10^2 	ext{ kg/m}^3$, and has a specific heat capacity of $2000 	ext{ J kg}^{-1}	ext{K}^{-1}$. Liquid B is denser at $10^3 	ext{ kg/m}^3$ and has a much higher specific heat capacity of $4000 	ext{ J kg}^{-1}	ext{K}^{-1}$.

The question asks us to predict their future. If we were to plot their temperatures over time, what would the graphs look like? Which one wins the race to room temperature?

### The Master Equation: Newton's Law of Cooling

To solve this, we need a mathematical tool that governs how things cool down. Enter **Newton's Law of Cooling**. This law states that the rate at which an object loses heat is directly proportional to the temperature difference between the object and its surroundings.

Mathematically, the rate of heat loss is given by:
$$ \frac{dQ}{dt} = hA(T - T_0) $$

Here, $h$ is the heat transfer coefficient (which depends on the emissivity and the surrounding medium), $A$ is the exposed surface area, $T$ is the temperature of the liquid, and $T_0$ is the ambient room temperature.

But we don't just want the rate of *heat* loss; we want the rate of *temperature* drop. We know from calorimetry that heat relates to temperature through mass and specific heat:
$$ dQ = -ms \, dT $$

Substituting this into our cooling equation gives:
$$ -ms \frac{dT}{dt} = hA(T - T_0) $$

Rearranging for the rate of cooling, we get:
$$ \frac{dT}{dt} = -\frac{hA}{ms}(T - T_0) $$

### Unpacking the Mass: Density and Volume

The problem doesn't give us the mass $m$ directly. Instead, it gives us the volume $V$ and the density $ho$. We know that mass is simply density times volume ($m = ho V$). Let's substitute this into our master equation:

$$ \frac{dT}{dt} = -\frac{hA}{ho V s}(T - T_0) $$

Now, let's look at the constants. Both beakers are identical, so they have the same surface area $A$ and the same volume $V$. The problem explicitly states they have the same emissivity, meaning $h$ is the same. They are in the same room, so $T_0$ is the same. And they both start at $60^{\circ}	ext{C}$.

This means that the entire expression $\frac{hA}{V}(T - T_0)$ is a constant for both beakers at any given temperature! The only variables that differ between beaker A and beaker B are the density $ho$ and the specific heat $s$.

Therefore, the magnitude of the rate of cooling is inversely proportional to the product of density and specific heat:
$$ \left| \frac{dT}{dt} ight| \propto \frac{1}{ho s} $$

This product, $ho s$, is known as the **volumetric heat capacity**. It represents the "thermal inertia" of the liquid per unit volume. A higher volumetric heat capacity means the liquid is more stubborn—it holds onto its temperature more tightly.

### The Final Calculation: Who Cools Faster?

Let's calculate this thermal inertia for both liquids.

For Liquid A:
$$ ho_A s_A = (8 	imes 10^2 	ext{ kg/m}^3) 	imes (2000 	ext{ J kg}^{-1}	ext{K}^{-1}) $$
$$ ho_A s_A = 16 	imes 10^5 	ext{ J/m}^3	ext{K} $$

For Liquid B:
$$ ho_B s_B = (10^3 	ext{ kg/m}^3) 	imes (4000 	ext{ J kg}^{-1}	ext{K}^{-1}) $$
$$ ho_B s_B = 40 	imes 10^5 	ext{ J/m}^3	ext{K} $$

Comparing the two, it is glaringly obvious that $ho_A s_A < ho_B s_B$.

Because the rate of cooling is *inversely* proportional to this product, the smaller thermal inertia of Liquid A means it has a **higher rate of cooling**. 
$$ \left| \frac{dT}{dt} ight|_A > \left| \frac{dT}{dt} ight|_B $$

Liquid A is losing its temperature faster than Liquid B.

### Visualizing the Race

If Liquid A cools faster, its temperature at any time $t > 0$ will be lower than Liquid B's temperature. On a graph of Temperature ($T$) versus time ($t$), the curve for A must plunge downward more steeply and lie **below** the curve for B.

Furthermore, look at the differential equation: $\frac{dT}{dt} \propto -(T - T_0)$. The rate of change of temperature depends on the temperature itself. This is the classic signature of an **exponential decay**, not a linear relationship. The graphs must be curves that asymptotically approach room temperature, not straight lines.

Looking at the options, we need the one where both graphs are exponential curves, and curve A is below curve B. This perfectly matches the schematic shown in option (b).

And there you have it! By breaking down the physics of heat transfer and understanding the concept of volumetric heat capacity, we've successfully predicted the thermal destiny of our two beakers.

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Visualized Solution

$Initial Setup$

$Newton’s Law of Cooling$

$Rate of Cooling$

$Thermal Inertia of A$

$Thermal Inertia of B$

$Comparing Rates$

$Graph Analysis$

$Food for Thought$

The Sigma Insight: Heat Transfer

The Setup

The Master Equation

Unpacking the Mass

The Final Calculation

Visualizing the Race

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Law of Cooling

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

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

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

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

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

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