20 mL of 80°C water is mixed with 20 mL of 0°C water in a perfect calorimeter. What is the final temperature?

To find the final temperature of the mixture, we can apply the principle of conservation of energy, which dictates that the heat lost by the hot water will be equal to the heat gained by the cold water.

Let's denote the final temperature as \( T_f \). The heat lost by the hot water is equal to the heat gained by the cold water.

Let's use the formula: \[ Q = mc\Delta T \] where: \( Q \) = heat energy (in joules, J) \( m \) = mass of the water (in kilograms, kg) \( c \) = specific heat capacity of the water (4.18 J/g°C)

Since we are working in milliliters and degrees Celsius and the densities of water at the temperatures given are almost identical, we can simplify by stating that the mass of water in grams is numerically equivalent to the volume in milliliters. And since the volumes and masses of the two amounts of water are equal, we can set the mass of both to be the same, and it will cancel out in our equation.

The heat lost by the hot water is: \[ Q_{hot} = m_{hot} \cdot c_{water} \cdot (T_{hot} - T_f) \]

The heat gained by the cold water is: \[ Q_{cold} = m_{cold} \cdot c_{water} \cdot (T_f - T_{cold}) \]

We know that \( Q_{hot} = Q_{cold} \), so:

\[ m_{hot} \cdot c_{water} \cdot (T_{hot} - T_f) = m_{cold} \cdot c_{water} \cdot (T_f - T_{cold}) \]

With \( m_{hot} = m_{cold} \) (since they're both 20 mL of water), and \( c_{water} \) is a constant that cancels out from both sides, we get:

\[ (T_{hot} - T_f) = (T_f - T_{cold}) \]

Plugging in \( T_{hot} = 80°C \) and \( T_{cold} = 0°C \), we have:

\[ (80°C - T_f) = (T_f - 0°C) \] \[ 80°C - T_f = T_f \] \[ 80°C = 2T_f \] \[ T_f = 40°C \]

Therefore, the final temperature of the mixture in the perfect calorimeter will be 40°C.