Animated Solution for Physics - Properties of Solids and Liquids: Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is

A Tale of Two Metals

Imagine you are in a laboratory, conducting a classic thermodynamics experiment. You have two distinct metal bodies, $A$ and $B$. They are perfectly matched in mass, so $m_A=m_B=m$. You place them on identical heaters that supply heat at a perfectly uniform rate.

Physically, this means the power input is constant. The rate at which heat energy is pumped into the bodies, denoted as $\frac{dQ}{dt}$, is identical for both. As they absorb this heat, their temperatures begin to rise, tracing out the beautiful linear graphs we see in the problem.

Calorimetry in Motion

To decode the graph, we need to bridge the gap between the heat supplied and the temperature change. We start with the fundamental equation of calorimetry:

Since we are dealing with a continuous process over time, we differentiate this equation with respect to time $t$:

Here, $\frac{dT}{dt}$ is the rate of change of temperature, which geometrically represents the slope of the Temperature-Time ($T-t$) graph. Because both bodies are heated under similar conditions, their heat supply rates are equal:

Substituting our differentiated equation into this equality gives:

The identical masses ($m$) gracefully cancel out. Rearranging the terms to isolate the ratio of specific heat capacities, we uncover a profound inverse relationship:

This tells us that the specific heat capacity is inversely proportional to the slope of the heating curve.

The Slopes of A and B

Now, we turn our attention to the visual data. We need to extract the slopes for both lines from the graph.

For body $A$, the line starts at the origin $(0,0)$ and passes cleanly through the grid intersection at $t=3\text{ s}$ and $T=120^{\circ }\text{C}$. The slope is simply the rise over run:

Similarly, for body $B$, the line starts at the origin and passes through $t=6\text{ s}$ and $T=90^{\circ }\text{C}$. Calculating its slope yields:

Calculating the Ratio

With our slopes in hand, the final calculation is a breeze. We substitute these values back into our derived ratio equation:

Dividing both the numerator and the denominator by their greatest common divisor, $5$, we arrive at the final, elegant fraction:

Why the Steeper Line Means Less Specific Heat

Before we wrap up, let's pause and appreciate the physical reality behind the math. Look at the graph again. Line $A$ is much steeper than line $B$. This means body $A$'s temperature skyrockets quickly when heat is applied.

Why does this happen? Because body $A$ has a lower specific heat capacity. It requires very little thermal energy to raise its temperature. Conversely, body $B$ is sluggish; its temperature rises slowly because it has a higher specific heat capacity, acting like a thermal sponge that absorbs a lot of heat before showing a temperature change. The math perfectly mirrors the physics!