In electrical calorimeter experiment, voltage across the heater is `100.0 V` and current is `10.0 A`. Heater is switched on for t `= 700.0 s`. Room temperature is `theta_0 = 10.0^(@)C` and final temperature of calorimeter and unknown liquid is `theta_f = 73.0^(@) C`. Mass of empty calorimeter is `m_1 = 1.0kg` and combined mass of calorimeter and unknown liquid is `m_2 = 3.0 kg`. Find the specificheat capacity of the unknown liquid in proper significant figures. Specific heat of calorimerter = `3.0xx10^3 j//kg .^(@) C`.

Explanation

To find the specific heat capacity of the unknown liquid in the electrical calorimeter experiment, we can follow these steps:

### Step 1: Write down the given data
- Voltage across the heater, \( V = 100.0 \, \text{V} \)
- Current through the heater, \( I = 10.0 \, \text{A} \)
- Time the heater is switched on, \( t = 700.0 \, \text{s} \)
- Initial temperature of the calorimeter, \( \theta_0 = 10.0 \, ^\circ \text{C} \)
- Final temperature of the calorimeter and unknown liquid, \( \theta_f = 73.0 \, ^\circ \text{C} \)
- Mass of the empty calorimeter, \( m_1 = 1.0 \, \text{kg} \)
- Combined mass of the calorimeter and unknown liquid, \( m_2 = 3.0 \, \text{kg} \)
- Specific heat of the calorimeter, \( C_c = 3.0 \times 10^3 \, \text{J/(kg} \cdot ^\circ \text{C)} \)

### Step 2: Calculate the heat supplied to the system
Using the formula for electrical power:
\[ H = V \cdot I \cdot t \]
Substituting the values:
\[ H = 100.0 \, \text{V} \cdot 10.0 \, \text{A} \cdot 700.0 \, \text{s} = 700000 \, \text{J} \]

### Step 3: Calculate the change in temperature
The change in temperature (\( \Delta \theta \)) is given by:
\[ \Delta \theta = \theta_f - \theta_0 = 73.0 \, ^\circ \text{C} - 10.0 \, ^\circ \text{C} = 63.0 \, ^\circ \text{C} \]

### Step 4: Calculate the mass of the unknown liquid
The mass of the unknown liquid (\( m_l \)) can be calculated as:
\[ m_l = m_2 - m_1 = 3.0 \, \text{kg} - 1.0 \, \text{kg} = 2.0 \, \text{kg} \]

### Step 5: Set up the heat balance equation
The heat gained by the calorimeter and the unknown liquid is equal to the heat supplied:
\[ H = m_1 C_c \Delta \theta + m_l S_l \Delta \theta \]
Where \( S_l \) is the specific heat capacity of the unknown liquid. Rearranging gives:
\[ S_l = \frac{H - m_1 C_c \Delta \theta}{m_l \Delta \theta} \]

### Step 6: Substitute the known values
Substituting the values into the equation:
\[ S_l = \frac{700000 \, \text{J} - (1.0 \, \text{kg} \cdot 3.0 \times 10^3 \, \text{J/(kg} \cdot ^\circ \text{C)} \cdot 63.0 \, ^\circ \text{C})}{2.0 \, \text{kg} \cdot 63.0 \, ^\circ \text{C}} \]

Calculating the heat absorbed by the calorimeter:
\[ = 1.0 \cdot 3.0 \times 10^3 \cdot 63.0 = 189000 \, \text{J} \]

Now substituting this back into the equation:
\[ S_l = \frac{700000 \, \text{J} - 189000 \, \text{J}}{2.0 \cdot 63.0} \]
\[ = \frac{511000 \, \text{J}}{126.0} \approx 4055.56 \, \text{J/(kg} \cdot ^\circ \text{C)} \]

### Step 7: Round to proper significant figures
The specific heat capacity of the unknown liquid is:
\[ S_l \approx 4.1 \times 10^3 \, \text{J/(kg} \cdot ^\circ \text{C)} \]