Calculate the temperature found when 10kg of water at 100°C is mixed with 20kg of water at 40°C.
heat lost by warm water + heat gained by cool water = 0
[mass warm water x specific heat water x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
From the formula displaced , the T final is : H loss by warm H2O + H gained by cold H2O =0. M C ( T2-T1) + M C ( T2-T1 ) =0 .T final= ... Continue...
answer
To calculate the final temperature when two substances are mixed together, we need to consider the principle of conservation of energy. The energy lost by one substance (in this case, water at a higher temperature) is equal to the energy gained by the other substance (water at a lower temperature) when they reach thermal equilibrium.
To solve this problem, we can use the principle of conservation of energy, specifically the equation:
\(Q_{\text{{lost}}} = Q_{\text{{gained}}}\)
where \(Q_{\text{{lost}}}\) represents the energy lost by the hot water and \(Q_{\text{{gained}}}\) represents the energy gained by the cold water.
To calculate the energy lost or gained, we use the formula:
\(Q = mc\Delta T\)
where:
- \(Q\) is the heat energy in Joules (J)
- \(m\) is the mass of the substance in kilograms (kg)
- \(c\) is the specific heat capacity of the substance in J/(kg·°C)
- \(\Delta T\) is the change in temperature in Celsius (°C)
For the hot water at 100°C, let's assume its specific heat capacity is \(c_1 = 4.18\) J/(g·°C). Since the mass is given as 10 kg, we need to convert it to grams:
\(m_1 = 10,000\) g
For the cold water at 40°C, let's assume its specific heat capacity is \(c_2 = 4.18\) J/(g·°C). Since the mass is given as 20 kg, we need to convert it to grams:
\(m_2 = 20,000\) g
Now we can calculate the energy lost and gained:
\(Q_{\text{{lost}}} = m_1c_1\Delta T_1\)
where \(\Delta T_1 = 100°C - T_\text{{final}}\)
\(Q_{\text{{gained}}} = m_2c_2\Delta T_2\)
where \(\Delta T_2 = T_\text{{final}} - 40°C\)
Since the total energy lost is equal to the total energy gained, we can set:
\(m_1c_1\Delta T_1 = m_2c_2\Delta T_2\)
Now we can substitute the values and solve for \(T_\text{{final}}\).