A system `S` receives heat continuously from an electric heater of power `10 W`. The temperature of `S` becomes constant at `50^(@)C` when the surrounding temperature is `20^(@)C`. After the heater is switched off, `S` cools from `35.1^(@)C` to `34.9^(@)C` in `1 minute`. the heat capacity of `S` is

Options

A. `100 J//^(@)C`
B. `300 J//^(@)C`
C. `750 J//^(@)C`
D. `1500 J//^(@)C`

Explanation

To find the heat capacity of the system \( S \), we can follow these steps:

### Step 1: Understand the given information
- The power of the electric heater is \( P = 10 \, \text{W} \).
- The steady temperature of the system \( S \) is \( T_S = 50^\circ C \).
- The surrounding temperature is \( T_{sur} = 20^\circ C \).
- After the heater is switched off, the system cools from \( 35.1^\circ C \) to \( 34.9^\circ C \) in \( 1 \, \text{minute} \).

### Step 2: Calculate the heat loss rate using Newton's Law of Cooling
According to Newton's Law of Cooling, the heat loss rate \( \frac{dQ}{dt} \) can be expressed as:
\[ \frac{dQ}{dt} = k (T_{mean} - T_{sur}) \]
Where:
- \( T_{mean} \) is the mean temperature of the system and the surrounding.
- \( k \) is the cooling constant.

### Step 3: Calculate the mean temperature
The mean temperature can be calculated as:
\[ T_{mean} = \frac{T_S + T_{sur}}{2} = \frac{50 + 20}{2} = 35^\circ C \]

### Step 4: Set up the equation for heat loss rate
Since the power of the heater is equal to the heat loss rate when the system is at a steady temperature:
\[ 10 \, \text{W} = k (35 - 20) \]
\[ 10 = k \cdot 15 \]
Solving for \( k \):
\[ k = \frac{10}{15} = \frac{2}{3} \, \text{W/°C} \]

### Step 5: Calculate the heat loss when cooling from \( 35.1^\circ C \) to \( 34.9^\circ C \)
The temperature difference \( \Delta T \) is:
\[ \Delta T = 35.1 - 34.9 = 0.2 \, \text{°C} \]
The time interval is \( t = 1 \, \text{minute} = 60 \, \text{seconds} \).

Using the heat loss formula:
\[ \frac{dQ}{dt} = k (T_{mean} - T_{sur}) = \frac{2}{3} (34.9 - 20) \]
Calculating \( T_{mean} \) for the cooling phase:
\[ T_{mean} = \frac{35.1 + 34.9}{2} = 35 \, \text{°C} \]
So,
\[ \frac{dQ}{dt} = \frac{2}{3} (35 - 20) = \frac{2}{3} \cdot 15 = 10 \, \text{W} \]

### Step 6: Calculate total heat lost over 1 minute
The total heat lost \( Q \) over 1 minute is:
\[ Q = \frac{dQ}{dt} \cdot t = 10 \, \text{W} \cdot 60 \, \text{s} = 600 \, \text{J} \]

### Step 7: Relate heat loss to heat capacity
Using the relation:
\[ Q = C \cdot \Delta T \]
Where \( C \) is the heat capacity. Rearranging gives:
\[ C = \frac{Q}{\Delta T} = \frac{600 \, \text{J}}{0.2 \, \text{°C}} = 3000 \, \text{J/°C} \]

### Final Answer
The heat capacity of the system \( S \) is \( C = 3000 \, \text{J/°C} \).

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