To find the final temperature when mixing different masses and temperatures of water, we can use the principle of conservation of energy. The heat lost by the hotter water will equal the heat gained by the cooler water until thermal equilibrium is reached.
Let’s define:
- \( m_1 = 3 , \text{kg} \) (the mass of water at 20°C)
- \( T_1 = 20 , \text{°C} \) (the initial temperature of water \( m_1 \))
- \( m_2 = 10 , \text{kg} \) (the mass of water at 60°C)
- \( T_2 = 60 , \text{°C} \) (the initial temperature of water \( m_2 \))
- \( T_f \) = final temperature of the mixture
We can set up the equation for heat lost and gained: \[ m_2 \cdot C \cdot (T_2 - T_f) = m_1 \cdot C \cdot (T_f - T_1) \]
Where \( C \) is the specific heat of water, which cancels out from both sides: \[ 10 \cdot (60 - T_f) = 3 \cdot (T_f - 20) \]
Expanding both sides: \[ 600 - 10T_f = 3T_f - 60 \]
Rearranging the equation to isolate \( T_f \): \[ 600 + 60 = 10T_f + 3T_f \] \[ 660 = 13T_f \]
Now, solving for \( T_f \): \[ T_f = \frac{660}{13} \approx 50.77 , \text{°C} \]
Thus, the final temperature of the mixture will be approximately 50.77°C.