An immersion heater with electrical resistance `7Omega` is immersed in 0.1 kg of water at `20^@ C` for 3 min. If the flow of current is 4 A , what is the final temperature of the water assuming whole of heat produced is consumed in water. (Specific heat capacity of water = `4.2 xx 10^(3) J kg^(-1)K^(-1)`)

Options

A. `28^@C`
B. `48@C`
C. `52^@C`
D. `68^@C`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat produced by the immersion heater.
The heat produced by the immersion heater can be calculated using the formula:
\[ Q = I^2 R T \]
Where:
- \( Q \) = heat produced (in Joules)
- \( I \) = current (in Amperes)
- \( R \) = resistance (in Ohms)
- \( T \) = time (in seconds)

Given:
- \( I = 4 \, \text{A} \)
- \( R = 7 \, \Omega \)
- \( T = 3 \, \text{min} = 3 \times 60 = 180 \, \text{s} \)

Substituting the values:
\[ Q = (4)^2 \times 7 \times 180 \]
\[ Q = 16 \times 7 \times 180 \]
\[ Q = 112 \times 180 \]
\[ Q = 20160 \, \text{J} \]

### Step 2: Calculate the heat gained by the water.
The heat gained by the water can be calculated using the formula:
\[ Q = mc\Delta T \]
Where:
- \( m \) = mass of the water (in kg)
- \( c \) = specific heat capacity of water (in J/kg/K)
- \( \Delta T \) = change in temperature (in K or °C)

Given:
- \( m = 0.1 \, \text{kg} \)
- \( c = 4.2 \times 10^3 \, \text{J/kg/K} \)
- Initial temperature \( T_i = 20 \, \text{°C} \)

Let the final temperature be \( T_f \). The change in temperature \( \Delta T = T_f - T_i = T_f - 20 \).

Substituting into the heat gained equation:
\[ Q = mc(T_f - T_i) \]
\[ 20160 = 0.1 \times 4.2 \times 10^3 \times (T_f - 20) \]

### Step 3: Solve for the final temperature \( T_f \).
First, simplify the equation:
\[ 20160 = 420 \times (T_f - 20) \]

Now, divide both sides by 420:
\[ \frac{20160}{420} = T_f - 20 \]
\[ 48 = T_f - 20 \]

Now, add 20 to both sides:
\[ T_f = 48 + 20 \]
\[ T_f = 68 \, \text{°C} \]

### Final Answer:
The final temperature of the water is \( 68 \, \text{°C} \).

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