An electric heater of power 100 W raises the temperature of 5 kg of a liquid from `25^(@)C` to `31^(@)C` in 2 minutes. Calculate the specific heat of the liquid.

Answer

To calculate the specific heat of the liquid, we can follow these steps:

### Step 1: Identify the given values
- Power of the heater (P) = 100 W
- Mass of the liquid (m) = 5 kg
- Initial temperature (T_initial) = 25°C
- Final temperature (T_final) = 31°C
- Time (t) = 2 minutes = 2 × 60 seconds = 120 seconds

### Step 2: Calculate the change in temperature (ΔT)
\[ \Delta T = T_{final} - T_{initial} = 31°C - 25°C = 6°C \]

### Step 3: Calculate the heat generated by the heater (Q)
The heat generated by the heater can be calculated using the formula:
\[ Q = P \times t \]
Substituting the known values:
\[ Q = 100 \, \text{W} \times 120 \, \text{s} = 12000 \, \text{J} \]

### Step 4: Use the formula for heat gained by the liquid
The heat gained by the liquid can be expressed as:
\[ Q = mc\Delta T \]
Where:
- \( c \) is the specific heat capacity of the liquid.

### Step 5: Rearrange the formula to solve for specific heat (c)
\[ c = \frac{Q}{m \Delta T} \]
Substituting the known values:
\[ c = \frac{12000 \, \text{J}}{5 \, \text{kg} \times 6°C} \]

### Step 6: Calculate the specific heat (c)
\[ c = \frac{12000 \, \text{J}}{30 \, \text{kg°C}} = 400 \, \text{J/(kg°C)} \]

### Final Answer
The specific heat of the liquid is \( 400 \, \text{J/(kg°C)} \).

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