A tap supplies water at `22^(@)C`, a man takes 1 L of water per min at `37^(@)C` from the geyser. The power of geyser is

Options

A. 525 W
B. 1050 W
C. 1775 W
D. 2100 W

Answer

To find the power of the geyser, we can follow these steps:

### Step 1: Identify the temperatures
The water is supplied at a temperature of \( T_1 = 22^\circ C \) and heated to \( T_2 = 37^\circ C \).

### Step 2: Calculate the temperature difference
The temperature difference \( \Delta T \) is calculated as:
\[ \Delta T = T_2 - T_1 = 37^\circ C - 22^\circ C = 15^\circ C \]

### Step 3: Determine the mass flow rate of water
The man takes 1 liter of water per minute. We need to convert this to a mass flow rate in kilograms per second. Since the density of water is approximately \( 1 \, \text{kg/L} \):
\[ m = 1 \, \text{L/min} = \frac{1 \, \text{kg}}{60 \, \text{s}} = \frac{1}{60} \, \text{kg/s} \]

### Step 4: Use the specific heat of water
The specific heat capacity \( S \) of water is given as:
\[ S = 4200 \, \text{J/(kg} \cdot \text{K)} \]

### Step 5: Calculate the heat energy required
The heat energy \( Q \) required to heat the water can be calculated using the formula:
\[ Q = m \cdot S \cdot \Delta T \]
Substituting the values:
\[ Q = \left(\frac{1}{60} \, \text{kg/s}\right) \cdot (4200 \, \text{J/(kg} \cdot \text{K)}) \cdot (15 \, \text{K}) \]

### Step 6: Calculate the power of the geyser
Power \( P \) is defined as the rate of heat transfer, which is equal to \( Q \):
\[ P = Q = \left(\frac{1}{60}\right) \cdot 4200 \cdot 15 \]
Calculating this gives:
\[ P = \frac{1}{60} \cdot 4200 \cdot 15 = 1050 \, \text{W} \]

### Conclusion
Thus, the power of the geyser is:
\[ P = 1050 \, \text{W} \]