A 10 W electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and container rises by `3^(@)`K in 15 min. The container is then emptied, dired and filled with 2kg of oil. The same heater now raises the temperature of container oil system by 2K in 20 min. Assume there is no heat loss in the process and the specific heat of water is 4200`Jkg^(-1)K^(-1)`, the specific heat of oil in the same limit is equal to

Options

A. `1.50 xx 10^(3)`
B. `2.55 xx 10^(3)`
C. `3.00 xx 10^(3)`
D. `2.10 xx 10^(3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat supplied by the heater when heating water
The power of the heater is given as 10 W. The time for which the heater is used is 15 minutes, which we need to convert to seconds:
\[ \text{Time} = 15 \text{ minutes} = 15 \times 60 = 900 \text{ seconds} \]
The heat supplied by the heater can be calculated using the formula:
\[ Q = P \times t \]
Where \( P \) is the power and \( t \) is the time. Thus,
\[ Q = 10 \text{ W} \times 900 \text{ s} = 9000 \text{ J} \]

### Step 2: Calculate the heat absorbed by the water
The heat absorbed by the water can be calculated using the formula:
\[ Q = mc\Delta T \]
Where:
- \( m \) is the mass of the water (0.5 kg),
- \( c \) is the specific heat of water (4200 J/kg·K),
- \( \Delta T \) is the temperature rise (3 K).

Substituting the values:
\[ Q_{\text{water}} = 0.5 \text{ kg} \times 4200 \text{ J/kg·K} \times 3 \text{ K} = 6300 \text{ J} \]

### Step 3: Calculate the heat absorbed by the container
Let \( C \) be the thermal capacity of the container. The heat absorbed by the container can be expressed as:
\[ Q_{\text{container}} = C \Delta T \]
Where \( \Delta T = 3 \text{ K} \). Therefore, we have:
\[ Q_{\text{container}} = C \times 3 \text{ K} \]

### Step 4: Set up the equation for the total heat supplied
The total heat supplied by the heater is equal to the sum of the heat absorbed by the water and the container:
\[ 9000 \text{ J} = 6300 \text{ J} + C \times 3 \text{ K} \]
Rearranging gives:
\[ C \times 3 = 9000 - 6300 = 2700 \text{ J} \]
Thus,
\[ C = \frac{2700}{3} = 900 \text{ J/K} \]

### Step 5: Calculate the heat supplied when heating oil
Now, the container is filled with 2 kg of oil, and the heater raises the temperature by 2 K in 20 minutes. First, convert 20 minutes to seconds:
\[ \text{Time} = 20 \text{ minutes} = 20 \times 60 = 1200 \text{ seconds} \]
The heat supplied by the heater is:
\[ Q = 10 \text{ W} \times 1200 \text{ s} = 12000 \text{ J} \]

### Step 6: Calculate the heat absorbed by the oil
Let \( c_o \) be the specific heat of the oil. The heat absorbed by the oil is:
\[ Q_{\text{oil}} = m c_o \Delta T \]
Where:
- \( m = 2 \text{ kg} \),
- \( \Delta T = 2 \text{ K} \).
Thus,
\[ Q_{\text{oil}} = 2 \text{ kg} \times c_o \times 2 \text{ K} = 4c_o \text{ J} \]

### Step 7: Calculate the heat absorbed by the container when heating oil
The heat absorbed by the container when heating the oil is:
\[ Q_{\text{container}} = C \Delta T = 900 \text{ J/K} \times 2 \text{ K} = 1800 \text{ J} \]

### Step 8: Set up the equation for the total heat supplied when heating oil
The total heat supplied by the heater is equal to the sum of the heat absorbed by the oil and the container:
\[ 12000 \text{ J} = 4c_o + 1800 \text{ J} \]
Rearranging gives:
\[ 4c_o = 12000 - 1800 = 10200 \text{ J} \]
Thus,
\[ c_o = \frac{10200}{4} = 2550 \text{ J/kg·K} \]

### Final Answer
The specific heat of the oil is:
\[ c_o = 2550 \text{ J/kg·K} \]