A 10.0 W electrical heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of the water and the container rise by 3 K in 15 minutes. The container is then emptied, dried, and filled with 2 kg of an oil. It is now observed that the same heater raises the temperature of the container-oil system by 2 K in 20 minutes. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is

Options

A. `2.5 xx 10^3 JK^(-1) kg^(-1)`
B. `5.1 xx 10^(3) JK^(-1) Kg^(-1)`
C. `3.0 xx 10^3 JK^(-1) Kg^(-1)`
D. `1.5 xx 10^(3) JK^(-1) kg^(-1)`

Explanation

To find the specific heat capacity of the oil, we can 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 it heats the water is 15 minutes. We need to convert this time into seconds:
\[ \text{Time} = 15 \text{ minutes} \times 60 \text{ seconds/minute} = 900 \text{ seconds} \]
The heat supplied \( Q \) by the heater can be calculated using the formula:
\[ Q = \text{Power} \times \text{Time} = 10 \text{ W} \times 900 \text{ s} = 9000 \text{ J} \]

### Step 2: Set up the equation for the water and container
The heat supplied to the water and the container is used to raise their temperature. The change in temperature is given as 3 K. Let \( m_c \) be the mass of the container and \( s_c \) be the specific heat capacity of the container. The equation for heat absorbed can be expressed as:
\[ Q = m_c s_c \Delta T + m_w s_w \Delta T \]
Where:
- \( m_w = 0.5 \text{ kg} \) (mass of water)
- \( s_w = 4200 \text{ J/kg·K} \) (specific heat capacity of water)
- \( \Delta T = 3 \text{ K} \)

Substituting the values:
\[ 9000 = m_c s_c (3) + (0.5)(4200)(3) \]
Calculating the heat absorbed by the water:
\[ 9000 = m_c s_c (3) + 6300 \]
Rearranging gives:
\[ m_c s_c (3) = 9000 - 6300 = 2700 \]
Thus:
\[ m_c s_c = \frac{2700}{3} = 900 \text{ J/K} \]

### Step 3: 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 the time:
\[ \text{Time} = 20 \text{ minutes} \times 60 \text{ seconds/minute} = 1200 \text{ seconds} \]
The heat supplied by the heater is:
\[ Q = 10 \text{ W} \times 1200 \text{ s} = 12000 \text{ J} \]

### Step 4: Set up the equation for the oil and container
Using the same heat equation:
\[ Q = m_c s_c (2) + m_o s_o (2) \]
Where:
- \( m_o = 2 \text{ kg} \) (mass of oil)
- \( s_o \) is the specific heat capacity of the oil.

Substituting the known values:
\[ 12000 = m_c s_c (2) + (2)(s_o)(2) \]
Substituting \( m_c s_c = 900 \):
\[ 12000 = 900(2) + 4s_o \]
Calculating:
\[ 12000 = 1800 + 4s_o \]
Rearranging gives:
\[ 4s_o = 12000 - 1800 = 10200 \]
Thus:
\[ s_o = \frac{10200}{4} = 2550 \text{ J/kg·K} \]

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