10 g of water at `20^(@)C` is mixed with 20 g of water at `10^(@)C` The resulting temperature is

Options

A. `15^(@)C`
B. `lt15^(@)C`
C. `gt15^(@)C`
D. `30^(@)C`

Answer

To find the resulting temperature when mixing two bodies of water at different temperatures, we can use the formula for the equilibrium temperature:

\[ T = \frac{M_1 \cdot T_1 + M_2 \cdot T_2}{M_1 + M_2} \]

Where:
- \( M_1 \) = mass of the first body of water
- \( T_1 \) = temperature of the first body of water
- \( M_2 \) = mass of the second body of water
- \( T_2 \) = temperature of the second body of water

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the first water (\( M_1 \)) = 10 g
- Temperature of the first water (\( T_1 \)) = \( 20^\circ C \)
- Mass of the second water (\( M_2 \)) = 20 g
- Temperature of the second water (\( T_2 \)) = \( 10^\circ C \)

2. **Substitute the values into the formula:**
\[ T = \frac{(10 \, \text{g} \cdot 20^\circ C) + (20 \, \text{g} \cdot 10^\circ C)}{10 \, \text{g} + 20 \, \text{g}} \]

3. **Calculate the numerator:**
- Calculate \( 10 \cdot 20 = 200 \)
- Calculate \( 20 \cdot 10 = 200 \)
- So, the numerator becomes \( 200 + 200 = 400 \)

4. **Calculate the denominator:**
- \( 10 + 20 = 30 \)

5. **Divide the numerator by the denominator:**
\[ T = \frac{400}{30} \approx 13.33^\circ C \]

6. **Conclusion:**
The resulting temperature when 10 g of water at \( 20^\circ C \) is mixed with 20 g of water at \( 10^\circ C \) is approximately \( 13.33^\circ C \).