Find the mass of water that vaporizes when 3.94 kg of mercury at 248 °C is added to 0.183 kg of water at 83.0 °C.
To determine the mass of water that vaporizes, we can use the equation:
Q = m * c * ΔT
Where:
Q - Heat energy absorbed or released
m - Mass of the substance
c - Specific heat capacity of the substance
ΔT - Change in temperature
First, let's find the heat energy absorbed or released by the water.
The specific heat capacity of water is 4.18 J/g°C.
Q_water = m_water * c_water * ΔT_water
Where:
m_water - Mass of water
c_water - Specific heat capacity of water
ΔT_water - Change in temperature of water
ΔT_water = T_final - T_initial
= (100 °C - 83.0 °C)
= 17 °C
Q_water = (0.183 kg) * (4.18 J/g°C) * (17 °C)
Next, let's find the heat energy needed for the mercury's temperature to reach 100 °C.
Q_mercury = m_mercury * c_mercury * ΔT_mercury
Where:
m_mercury - Mass of mercury
c_mercury - Specific heat capacity of mercury
ΔT_mercury - Change in temperature of mercury
ΔT_mercury = T_final - T_initial
= (100 °C - 248 °C)
= -148 °C (negative because the temperature decreases)
Q_mercury = (3.94 kg) * (0.14 J/g°C) * (-148 °C)
Next, since the heat released by mercury is absorbed by the water, we can equate the two equations:
Q_water = Q_mercury
(0.183 kg) * (4.18 J/g°C) * (17 °C) = (3.94 kg) * (0.14 J/g°C) * (-148 °C)
Now, we can solve for the mass of water that vaporizes.
Let's assume that the mass of water vaporized is represented as m_vaporized.
Q_mercury = Q_water
(3.94 kg) * (0.14 J/g°C) * (-148 °C) = (m_vaporized) * (40.7 J/g) * (100 °C)
Now we can solve for m_vaporized:
m_vaporized = [(3.94 kg) * (0.14 J/g°C) * (-148 °C)] / [(40.7 J/g) * (100 °C)]
m_vaporized ≈ -0.011 kg
Since mass cannot be negative, we can conclude that the mass of water vaporized is approximately 0.011 kg.