A metal cylinder of mass `0.5kg` is heated electically by a `12 W` heater in a room at `15^(@)C` The cylinder temperature rises uniformly to `25^(@)C` in 5 min and finally becomes constant at `45^(@)C` Asuming that the rate of heat loss is proportional to the excess temperature over the surroundings .

Options

A. the rate of loss of heat of the cylinder to surrounding at `20^@C` is 2 W
B. the rate of loss of heat of the cylinder to surrounding at `45^@C` is 12 W
C. the rate of loss of heat of the cylinder to surrounding at 20∘ C is 5W
D. the rate of loss of heat of the cylinder to surrounding at 45∘ C is 30W.

Answer

To solve the problem, we need to analyze the heating of the metal cylinder and the heat loss to the surroundings. We will follow these steps:

### Step 1: Understand the Given Information
- Mass of the cylinder, \( m = 0.5 \, \text{kg} \)
- Power of the heater, \( P = 12 \, \text{W} \)
- Initial temperature of the surroundings, \( T_0 = 15^\circ C \)
- Temperature after 5 minutes, \( T_1 = 25^\circ C \)
- Final constant temperature, \( T_f = 45^\circ C \)

### Step 2: Establish the Heat Loss Equation
According to the problem, the rate of heat loss \( \frac{dQ}{dt} \) is proportional to the excess temperature over the surroundings. This can be expressed as:
\[ \frac{dQ}{dt} = k (T - T_0) \]
where \( T \) is the temperature of the cylinder and \( k \) is a constant of proportionality.

### Step 3: Analyze the Final State
At the final constant temperature \( T_f = 45^\circ C \), the rate of heat loss equals the rate of heat supplied by the heater:
\[ k (T_f - T_0) = P \]
Substituting the values:
\[ k (45 - 15) = 12 \]
\[ k (30) = 12 \]
\[ k = \frac{12}{30} = \frac{2}{5} \, \text{W/}^\circ C \]

### Step 4: Calculate the Rate of Heat Loss at \( T_1 = 25^\circ C \)
Now we can find the rate of heat loss when the temperature of the cylinder is \( T_1 = 25^\circ C \):
\[ \frac{dQ}{dt} = k (T_1 - T_0) \]
Substituting the values:
\[ \frac{dQ}{dt} = \frac{2}{5} (25 - 15) \]
\[ \frac{dQ}{dt} = \frac{2}{5} (10) = 4 \, \text{W} \]

### Step 5: Calculate the Rate of Heat Loss at \( T_f = 45^\circ C \)
At the final temperature \( T_f = 45^\circ C \):
\[ \frac{dQ}{dt} = k (T_f - T_0) \]
Substituting the values:
\[ \frac{dQ}{dt} = \frac{2}{5} (45 - 15) \]
\[ \frac{dQ}{dt} = \frac{2}{5} (30) = 12 \, \text{W} \]

### Conclusion
From the calculations:
- The rate of heat loss at \( 25^\circ C \) is \( 4 \, \text{W} \).
- The rate of heat loss at \( 45^\circ C \) is \( 12 \, \text{W} \).

Thus, the correct statements based on the options provided are:
- The rate of loss of heat of the cylinder to the surrounding at \( 25^\circ C \) is \( 4 \, \text{W} \).
- The rate of loss of heat of the cylinder to the surrounding at \( 45^\circ C \) is \( 12 \, \text{W} \).