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`.
Assume that the temperature of the ball rises uniformly from `25^(@)C` to `39^(@)C "in" 2` minutes. Find the total loss of heat to the surrounding during this period.

Options

A. `900 J`
B. `940 J`
C. `960 J`
D. `1000J`

Explanation

To solve the problem step by step, we need to calculate the total loss of heat to the surroundings when the metal ball is heated from 25°C to 39°C over a period of 2 minutes.

### Step 1: Understand the Given Information
- Mass of the metal ball (m) = 2 kg
- Power of the heater (P) = 40 W
- Initial temperature of the ball (T_initial) = 25°C
- Final steady temperature of the ball (T_final) = 60°C
- Temperature after 2 minutes (T_intermediate) = 39°C
- Time (t) = 2 minutes = 2 × 60 seconds = 120 seconds

### Step 2: Calculate the Temperature Change
The temperature change (ΔT) from 25°C to 39°C is:
\[ \Delta T = T_{intermediate} - T_{initial} = 39°C - 25°C = 14°C \]

### Step 3: Calculate the Heat Supplied by the Heater
The heat supplied (Q_heater) by the heater can be calculated using the formula:
\[ Q_{heater} = P \times t \]
Substituting the values:
\[ Q_{heater} = 40 \, \text{W} \times 120 \, \text{s} = 4800 \, \text{J} \]

### Step 4: Calculate the Heat Absorbed by the Ball
The heat absorbed by the ball (Q_ball) can be calculated using the specific heat formula:
\[ Q_{ball} = m \cdot c \cdot \Delta T \]
However, we need to find the specific heat (c) first. Since we don't have the specific heat value, we will assume it is constant during the heating process.

Let's denote the specific heat of the ball as \( c \). The temperature change is 14°C, so:
\[ Q_{ball} = 2 \, \text{kg} \cdot c \cdot 14 \, \text{°C} = 28c \, \text{J} \]

### Step 5: Calculate the Total Loss of Heat to the Surroundings
The total loss of heat to the surroundings (Q_loss) can be calculated as:
\[ Q_{loss} = Q_{heater} - Q_{ball} \]
Substituting the values we have:
\[ Q_{loss} = 4800 \, \text{J} - 28c \, \text{J} \]

### Step 6: Conclusion
To find the exact value of \( Q_{loss} \), we need the specific heat \( c \) of the metal ball. However, we can express the total loss of heat to the surroundings as:
\[ Q_{loss} = 4800 - 28c \, \text{J} \]