Problem 85 - Master Resource Book in JEE Main Physics

Question

A solid cube and a solid sphere have equal surface areas. Both are at the same temperature of \(120^{\circ} \mathrm{C}\). Then (a) both of them will cool down at the same rate (b) the cube will cool down faster than the sphere (c) the sphere will cool down faster than the cube (d) whichever of the two is heavier will cool down faster

Step-by-Step Solution

Answer

The cube will cool down faster than the sphere (Option b).

Step 1: Calculate Cube Surface Area

The surface area of a cube with side length \(a\) is given by \(6a^2\). This formula represents the sum of the areas of the six identical square faces of the cube.

Step 2: Calculate Sphere Surface Area

The surface area of a sphere with radius \(r\) is given by \(4\pi r^2\). This formula represents the smooth continuous surface covering a sphere.

Step 3: Equate Surface Areas

Since the surfaces area of the cube and the sphere are equal, we equate the formula for the cube and the sphere: \[ 6a^2 = 4\pi r^2 \] to find a relationship between their geometric sizes.

Step 4: Express \(a\) in Terms of \(r\)

From the equation \(6a^2 = 4\pi r^2\), solve for \(a\): \[ a = \sqrt{\frac{2\pi}{3}} r \]. This gives the side length \(a\) of the cube in terms of the radius \(r\) of the sphere.

Step 5: Compare Volumes

The volume of the cube is \(a^3\) and for the sphere it is \(\frac{4}{3}\pi r^3\). Substitute \(a = \sqrt{\frac{2\pi}{3}} r\) into the cube’s volume and compare it with the sphere’s volume to determine their relative sizes.

Step 6: Calculate Cooling Rate

The rate at which an object cools depends on its mass (related to volume when density is constant). The volume of the sphere becomes \(\frac{4}{3}\pi r^3\) and the cube’s volume is \(( \sqrt{\frac{2\pi}{3}} r )^3 = \frac{2\pi}{3\sqrt{3}} r^3\). Since the sphere has a larger volume, it indicates that its mass is greater (assuming same density), and thus its relative cooling rate will be slower according to Newton's law of cooling.

Key Concepts

Cooling Rates

In the context of the cooling process, the rate at which an object cools can depend heavily on its mass and surface area. When two objects begin at the same initial temperature, the object that cools faster generally has a larger surface area to volume ratio. This is because a greater surface area allows more heat to be transferred away from the object, resulting in a faster rate of cooling.

However, mass is also an important factor to consider. A heavier object may cool relatively slower because it has more thermal energy stored within it, due to its larger mass. Thus, even if two objects have the same surface area, if one is significantly heavier, it will likely retain its heat longer and cool slower.

In our exercise, since the cube and sphere have equal surface areas but different volumes, their cooling rates will differ. The mass and volume of each shape can influence the outcome of the cooling process. The sphere ends up having a larger volume and mass than the cube, meaning it has more heat content and thus cools down slower than the cube.

Surface Area Comparison

Surface area is crucial in determining how heat is transferred between an object and its environment. To compare the cooling rates of a cube and a sphere with the same surface area, we first need to understand their geometric properties.

The surface area of a cube is calculated using the formula \(6a^2\), where \(a\) is the side length of the cube. For a sphere, the formula \(4\pi r^2\) is used, with \(r\) representing the radius.

In this exercise, the cube and the sphere have identical surface areas, so we equate \(6a^2 = 4\pi r^2\). Upon solving this equation, we derive the relationship between their dimensions: \(a = \sqrt{\frac{2\pi}{3}} r\).

Although they share the same surface area, their volumes differ. The cube’s volume decreases compared to its potential maximum size when the radius is fixed, while the sphere retains more volume with the same surface area. This difference in volume plays a key role in their cooling behaviors.

Newton's Law of Cooling

Newton's Law of Cooling describes how the rate of heat loss of a body is proportional to the difference in temperature between the body and its surroundings, assuming the temperature differential is not too large. The formula can be expressed as:

\[ \frac{dT}{dt} = -k(T - T_{ambient}) \]

where \(\frac{dT}{dt}\) represents the rate of change of temperature, \(T\) is the temperature of the object, \(T_{ambient}\) is the ambient temperature, and \(k\) is a positive constant dependent on the characteristics of the object and environment.

In the exercise, both the cube and the sphere start at the same temperature and share an identical surface area. According to Newton's Law, the cooling rate is influenced by the surface area, mass, and material properties. With equal surface areas, the difference in cooling will primarily arise from their masses. Since the sphere has a larger volume and therefore a larger mass, it will cool more slowly compared to the cube.

This principle helps us predict the cooling behavior of objects, which is particularly useful in real-world applications like engineering, cooking, and climate science.