80 g of water at `30^(@) C` is mixed with 50 g of water at `60^(@) C`, final temperature of mixture will be

Options

A. 30 g
B. 80 g
C. 1600 g
D. 150 g

Explanation

To find the final temperature of the mixture when 80 g of water at 30°C is mixed with 50 g of water at 60°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 temperatures:**
- Mass of water 1 (m1) = 80 g (at T1 = 30°C)
- Mass of water 2 (m2) = 50 g (at T2 = 60°C)

2. **Set up the heat transfer equation:**
The heat gained by the cooler water (m1) is equal to the heat lost by the warmer water (m2):
\[ m_1 \cdot C \cdot (T - T_1) = m_2 \cdot C \cdot (T_2 - T) \]
Here, C is the specific heat capacity of water, which cancels out since both masses are water.

3. **Substitute the known values:**
\[ 80 \cdot (T - 30) = 50 \cdot (60 - T) \]

4. **Expand both sides:**
\[ 80T - 2400 = 3000 - 50T \]

5. **Rearrange the equation:**
Combine like terms:
\[ 80T + 50T = 3000 + 2400 \]
\[ 130T = 5400 \]

6. **Solve for T:**
\[ T = \frac{5400}{130} \approx 41.54°C \]

### Final Answer:
The final temperature of the mixture is approximately **41.54°C**.