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

Explanation

To solve the problem step by step, we will follow the reasoning and calculations as outlined in the video transcript.

### Step 1: Understand the given information
- The power of the heater \( P = 10 \, \text{W} \)
- The steady temperature of the system \( T_s = 50^\circ C \)
- The surrounding temperature \( T_a = 20^\circ C \)
- The initial temperature after the heater is switched off \( T_i = 35.1^\circ C \)
- The final temperature after 1 minute \( T_f = 34.9^\circ C \)

### Step 2: Calculate the temperature difference
The temperature difference when the heater is on:
\[ \Delta T = T_s - T_a = 50^\circ C - 20^\circ C = 30^\circ C \]

### Step 3: Apply Newton's Law of Cooling
According to Newton's law of cooling, the rate of heat loss \( \frac{dQ}{dt} \) is proportional to the temperature difference between the system and the surroundings:
\[ \frac{dQ}{dt} = k \cdot (T - T_a) \]
Where \( k \) is a constant.

Given that \( \frac{dQ}{dt} = 10 \, \text{W} \) when the temperature of the system is \( 50^\circ C \):
\[ 10 = k \cdot (50 - 20) = k \cdot 30 \]
From this, we can solve for \( k \):
\[ k = \frac{10}{30} = \frac{1}{3} \, \text{W/°C} \]

### Step 4: Calculate the mean temperature after the heater is switched off
The mean temperature \( T_m \) when the heater is switched off can be calculated as:
\[ T_m = \frac{T_s + T_a}{2} = \frac{50 + 20}{2} = 35^\circ C \]

### Step 5: Calculate the heat loss in 1 minute
The heat loss \( dQ' \) over 1 minute (60 seconds) can be calculated using the formula:
\[ dQ' = k \cdot (T_m - T_a) \cdot t \]
Substituting the values:
\[ dQ' = \frac{1}{3} \cdot (35 - 20) \cdot 60 = \frac{1}{3} \cdot 15 \cdot 60 = \frac{900}{3} = 300 \, \text{J} \]

### Step 6: Calculate the change in temperature
The change in temperature \( \Delta T \) when the system cools from \( 35.1^\circ C \) to \( 34.9^\circ C \) is:
\[ \Delta T = T_i - T_f = 35.1 - 34.9 = 0.2^\circ C \]

### Step 7: Relate heat loss to heat capacity
Using the relationship:
\[ Q = S \cdot \Delta T \]
Where \( Q \) is the heat lost, \( S \) is the heat capacity, and \( \Delta T \) is the change in temperature. We can rearrange this to find \( S \):
\[ S = \frac{Q}{\Delta T} = \frac{300 \, \text{J}}{0.2^\circ C} = 1500 \, \text{J/°C} \]

### Final Answer
The heat capacity of the system \( S \) is:
\[ \boxed{1500 \, \text{J/°C}} \]