A resistance coil of 60 ohm is immersed in 42000 gm of water. A current of 7A is passed through it. Calculate the rise in temperature per minute. The sp. Heat of water = `4200(J)/(kg- ""^(@)K)`

Options

A. `5^(@)C`
B. `3^(@)C`
C. `2^(@)C`
D. `1^(@)C`

Answer

To solve the problem step by step, we will follow these calculations:

### Given Data:
- Resistance of the coil, \( R = 60 \, \Omega \)
- Mass of water, \( m = 42000 \, \text{g} = 42 \, \text{kg} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \))
- Current through the coil, \( I = 7 \, \text{A} \)
- Specific heat of water, \( s = 4200 \, \frac{J}{kg \cdot K} \)

### Step 1: Calculate the power (heat produced) by the resistance coil.
The power \( P \) dissipated in the resistor can be calculated using the formula:
\[ P = I^2 R \]
Substituting the values:
\[ P = (7 \, \text{A})^2 \times 60 \, \Omega = 49 \times 60 = 2940 \, \text{W} \]

### Step 2: Calculate the heat produced in one minute.
Since power is the rate of heat production, the total heat \( Q \) produced in one minute (60 seconds) is:
\[ Q = P \times t = 2940 \, \text{W} \times 60 \, \text{s} = 176400 \, \text{J} \]

### Step 3: Use the formula for heat transfer to find the rise in temperature.
The heat transfer formula is given by:
\[ Q = m s \Delta T \]
Where:
- \( Q \) is the heat added,
- \( m \) is the mass of the water,
- \( s \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

Rearranging the formula to find \( \Delta T \):
\[ \Delta T = \frac{Q}{m s} \]
Substituting the values:
\[ \Delta T = \frac{176400 \, \text{J}}{42 \, \text{kg} \times 4200 \, \frac{J}{kg \cdot K}} = \frac{176400}{176400} = 1 \, \text{K} \]

### Conclusion:
The rise in temperature per minute is:
\[ \Delta T = 1 \, \text{°C} \]

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