One litre of water at `30^@C` is mixed with one litre of water at `50^@C` . The temperature of the mixture will be

Options

A. `80^@C`
B. more than `50^@C` but less than `80^@C`
C. `20^@C`
D. between `30^@C` and `50^@C`

Explanation

To find the temperature of the mixture when 1 liter of water at 30°C is mixed with 1 liter of water at 50°C, we can use the principle of conservation of energy. The heat lost by the hotter water will be equal to the heat gained by the cooler water.

### Step-by-Step Solution:

1. **Identify the masses and specific heat capacities**:
- The mass of water at 30°C (m1) = 1 liter = 1 kg (since the density of water is 1 kg/L).
- The mass of water at 50°C (m2) = 1 liter = 1 kg.
- The specific heat capacity of water (c) = 4.18 kJ/kg°C (this value is not needed for calculation since it cancels out).

2. **Set up the equation for heat transfer**:
- Let T be the final temperature of the mixture.
- Heat gained by the cooler water (30°C):
\[ Q_1 = m_1 \cdot c \cdot (T - 30) \]
- Heat lost by the warmer water (50°C):
\[ Q_2 = m_2 \cdot c \cdot (50 - T) \]

3. **Apply the principle of conservation of energy**:
- According to the principle of conservation of energy:
\[ Q_1 = Q_2 \]
- Therefore, we can set up the equation:
\[ m_1 \cdot c \cdot (T - 30) = m_2 \cdot c \cdot (50 - T) \]
- Since both m1 and m2 are equal (1 kg) and c cancels out, we can simplify the equation to:
\[ (T - 30) = (50 - T) \]

4. **Solve for T**:
- Rearranging the equation:
\[ T - 30 = 50 - T \]
\[ T + T = 50 + 30 \]
\[ 2T = 80 \]
\[ T = 40 \]

5. **Conclusion**:
- The final temperature of the mixture is **40°C**.