To address your question about the cooling rates of two spherical balls, A and B, we need to clarify what we mean by the "rate of cooling." In this context, we are typically referring to the rate of loss of heat, which is influenced by the surface area and volume of the spheres, as well as the temperature difference between the balls and their surroundings.

Understanding the Cooling Process

When an object cools down, it loses heat to its surroundings. The rate at which it loses heat can be described by Newton's Law of Cooling, which states that the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings, provided this difference is not too large. Mathematically, this can be expressed as:

Q = hA(T - T_s)

Where:

• Q = rate of heat loss
• h = heat transfer coefficient
• A = surface area of the object
• T = temperature of the object
• T_s = temperature of the surroundings

Surface Area and Volume Considerations

For two spheres, the surface area (A) and volume (V) can be calculated using the following formulas:

Surface Area (A) = 4πr²

Volume (V) = (4/3)πr³

Let’s denote the radius of ball B as r. Then, the radius of ball A, which is twice the diameter of ball B, will be 2r. Now we can calculate the surface areas:

• Surface Area of B: A_B = 4πr²
• Surface Area of A: A_A = 4π(2r)² = 16πr²

Calculating the Ratio of Cooling Rates

Now, let’s look at the ratio of the rates of heat loss for the two balls:

Rate of heat loss for A (Q_A) = hA_A(T - T_s) = h(16πr²)(T - T_s)

Rate of heat loss for B (Q_B) = hA_B(T - T_s) = h(4πr²)(T - T_s)

Now, we can find the ratio of the rates of heat loss:

Ratio (Q_A : Q_B) = (h(16πr²)(T - T_s)) : (h(4πr²)(T - T_s))

After simplifying, we find:

Ratio (Q_A : Q_B) = 16 : 4 = 4 : 1

Conclusion on Cooling Rates

Thus, the ratio of the rates of cooling (in terms of heat loss) of balls A and B is 4:1. This means that ball A, being larger, will lose heat at a rate four times that of ball B, assuming they are both at the same initial temperature and in identical surroundings. This illustrates how size and surface area significantly affect the cooling process of objects.