A metal ball of mass `2kg` is heated means of a `40W` heater in a room at `25^(@)C`. The temperature of the ball beomes steady at `60^(@)C`.
Find the rate of loss of heat to the surrounding when the ball is at `60^(@)C`.

Options

A. `40 W`
B. `16 W`
C. `96 W`
D. `100W`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We have a metal ball of mass 2 kg that is heated by a 40W heater. The ball reaches a steady temperature of 60°C in a room at 25°C. We need to find the rate of heat loss to the surroundings when the ball is at 60°C.

### Step 2: Identify the Variables
- Mass of the ball (m) = 2 kg (not directly needed for heat loss calculation)
- Power of the heater (P) = 40 W
- Temperature of the ball (T1) = 60°C
- Temperature of the surroundings (T2) = 25°C

### Step 3: Use the Heat Transfer Formula
The rate of heat flow (dq/dt) can be expressed as:
\[ \frac{dq}{dt} = k (T_1 - T_2) \]
Where:
- \( k \) is a constant of proportionality,
- \( T_1 \) is the temperature of the ball,
- \( T_2 \) is the temperature of the surroundings.

### Step 4: Relate Power to Heat Flow
Since the heater is providing a constant power of 40 W, we can equate this to the heat flow when the ball is at 60°C:
\[ P = \frac{dq}{dt} = k (T_1 - T_2) \]
Substituting the known values:
\[ 40 = k (60 - 25) \]
\[ 40 = k (35) \]

### Step 5: Solve for Constant \( k \)
To find \( k \):
\[ k = \frac{40}{35} = \frac{8}{7} \, \text{W/°C} \]

### Step 6: Calculate Heat Loss at 60°C
Now, when the temperature of the ball is at 60°C, we need to find the rate of heat loss when the temperature drops to 39°C. The new temperature of the ball (T1) is now 39°C:
\[ \frac{dq}{dt} = k (T_1 - T_2) \]
Substituting the new values:
\[ \frac{dq}{dt} = \frac{8}{7} (39 - 25) \]
\[ \frac{dq}{dt} = \frac{8}{7} (14) \]
\[ \frac{dq}{dt} = \frac{112}{7} = 16 \, \text{W} \]

### Step 7: Conclusion
The rate of heat loss to the surroundings when the ball is at 60°C is **16 W**.

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