A 2100 W continuous flow geyser (instant geyser) has water inlet temperature = `10°C` while the water flows out at the rate of 20 g/sec. The outlet temperature of water must be about

Options

A. `20^(@)C`
B. `30^(@)C`
C. `35^(@)C`
D. `40^(@)C`

Answer

To find the outlet temperature of the water from the geyser, we can use the principle of conservation of energy, specifically focusing on the heat transfer involved in heating the water.

### Step-by-step Solution:

1. **Identify Given Data:**
- Power of the geyser, \( P = 2100 \, \text{W} \)
- Inlet temperature of water, \( T_{\text{in}} = 10^\circ \text{C} \)
- Mass flow rate of water, \( \dot{m} = 20 \, \text{g/s} = 0.02 \, \text{kg/s} \) (converted to kg)
- Specific heat capacity of water, \( s = 4200 \, \text{J/(kg} \cdot \text{K)} \)

2. **Convert Power to Heat Transfer Rate:**
- The power of the geyser indicates the rate of heat transfer to the water, which is \( Q = 2100 \, \text{J/s} \).

3. **Use the Formula for Heat Transfer:**
- The heat transfer to the water can be expressed as:
\[ Q = \dot{m} \cdot s \cdot \Delta T \]
- Where \( \Delta T \) is the change in temperature of the water.

4. **Rearranging the Formula to Find \( \Delta T \):**
- Rearranging gives us:
\[ \Delta T = \frac{Q}{\dot{m} \cdot s} \]

5. **Substituting the Values:**
- Substitute the known values into the equation:
\[ \Delta T = \frac{2100 \, \text{J/s}}{0.02 \, \text{kg/s} \cdot 4200 \, \text{J/(kg} \cdot \text{K)}} \]
- Calculate the denominator:
\[ 0.02 \cdot 4200 = 84 \, \text{J/K} \]
- Now calculate \( \Delta T \):
\[ \Delta T = \frac{2100}{84} \approx 25 \, \text{K} \]

6. **Calculate the Outlet Temperature:**
- The outlet temperature \( T_{\text{out}} \) can be found using:
\[ T_{\text{out}} = T_{\text{in}} + \Delta T \]
- Substitute the values:
\[ T_{\text{out}} = 10^\circ \text{C} + 25^\circ \text{C} = 35^\circ \text{C} \]

### Final Answer:
The outlet temperature of the water is approximately \( 35^\circ \text{C} \).

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