A 53.31-g sample of water at 88.8 degrees C is added to a sample of water at 23.9 degrees C in a constant- pressure calorimeter. If the final temperature of the combined water is 43.3 degrees C and the heat capacity of the calorimeter is 26.3 J/degrees C, calculate the mass of the water originally in the calorimeter.
[mass warm water x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal x (Tfinal-Tinitial)] = 0
To find the mass of the water originally in the calorimeter, we can use the principle of conservation of energy. The heat lost by the hot water is equal to the heat gained by the cold water and the calorimeter.
The equation to calculate heat is Q = m * C * ΔT, where Q is the heat, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat lost by the hot water using the equation Q_hot = m_hot * C_water * ΔT_hot.
We know the mass of the hot water (53.31 g), the specific heat capacity of water (4.18 J/g°C), and the change in temperature from the initial temperature (88.8°C) to the final temperature (43.3°C). So we can substitute these values into the equation:
Q_hot = (53.31 g) * (4.18 J/g°C) * (43.3°C - 88.8°C)
Next, let's calculate the heat gained by the cold water using the equation Q_cold = m_cold * C_water * ΔT_cold.
We don't know the mass of the cold water, so let's represent it as m_cold. The specific heat capacity of water is the same as before (4.18 J/g°C), and the change in temperature is from the initial temperature (23.9°C) to the final temperature (43.3°C). So we can set up the equation:
Q_cold = m_cold * (4.18 J/g°C) * (43.3°C - 23.9°C)
Lastly, let's calculate the heat gained by the calorimeter using the equation Q_calorimeter = C_calorimeter * ΔT_calorimeter.
We know the specific heat capacity of the calorimeter (26.3 J/°C) and the change in temperature is from the initial temperature (23.9°C) to the final temperature (43.3°C). So we have:
Q_calorimeter = (26.3 J/°C) * (43.3°C - 23.9°C)
According to the principle of conservation of energy, the heat lost (Q_hot) by the hot water is equal to the heat gained (Q_cold + Q_calorimeter) by the cold water and the calorimeter. Therefore, we can set up the equation:
Q_hot = Q_cold + Q_calorimeter
Now, we can plug in the calculated values and solve for the mass of the cold water (m_cold).
[(53.31 g) * (4.18 J/g°C) * (43.3°C - 88.8°C)] = [m_cold * (4.18 J/g°C) * (43.3°C - 23.9°C)] + [(26.3 J/°C) * (43.3°C - 23.9°C)]
Simplifying the equation will allow us to isolate m_cold, which gives us the mass of the water originally in the calorimeter.