A 35.0-g iron rod, initially at 25.6°C, is submerged into an unknown mass of water at 67.7°C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 63.5°C. What is the mass of the water?
[mass rod x specific heat rod x (Tfinal-Tintial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for the only unknown, mass H2O.
28.014 g
To solve this problem, we can use the principle of conservation of energy and the equation for heat transfer.
The heat gained by the iron rod is given by the formula:
Q_iron = m_iron * c_iron * ΔT_iron,
where m_iron is the mass of the iron rod, c_iron is the specific heat capacity of iron, and ΔT_iron is the change in temperature of the iron rod.
Similarly, the heat gained by the water is given by the formula:
Q_water = m_water * c_water * ΔT_water,
where m_water is the mass of water, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.
Since the container is insulated, the total heat gained by the rod must be equal to the total heat gained by the water:
Q_iron = Q_water.
We can rearrange the formula and express the mass of water:
m_water = (Q_iron) / (c_water * ΔT_water).
Given the data:
m_iron = 35.0 g,
c_iron = specific heat capacity of iron (which we can assume is 0.45 J/g°C),
ΔT_iron = final temperature - initial temperature of the iron = 63.5°C - 25.6°C = 37.9°C,
ΔT_water = final temperature - initial temperature of the water = 63.5°C - 67.7°C = -4.2°C (note that the change is negative because the water is losing heat).
Now we can calculate the mass of water:
m_water = (Q_iron) / (c_water * ΔT_water)
m_water = (m_iron * c_iron * ΔT_iron) / (c_water * ΔT_water).
Substituting the given values:
m_water = (35.0 g * 0.45 J/g°C * 37.9°C) / (4.18 J/g°C * -4.2°C)
Calculating this expression will give you the mass of water in grams.