A 32.3 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 62.9 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.4 ∘C.
To calculate the unknown mass of water in this problem, we can use the principle of conservation of heat. The heat lost by the iron rod equals the heat gained by the water.
First, we need to calculate the heat lost by the iron rod using the formula:
Q_iron = m_iron * c_iron * ΔT_iron
Where:
Q_iron is the heat lost by the iron rod,
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.
Given:
m_iron = 32.3 g
c_iron = 0.45 J/g°C (specific heat capacity of iron)
ΔT_iron = (final temperature - initial temperature) = (59.4 °C - 22.4 °C)
Substituting the given values, we can calculate the heat lost by the iron rod.
Q_iron = 32.3 g * 0.45 J/g°C * (59.4 °C - 22.4 °C)
Next, we can calculate the heat gained by the water using the formula:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water is the heat gained by the water,
m_water is the unknown mass of water,
c_water is the specific heat capacity of water, and
ΔT_water is the change in temperature of the water.
Given:
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = (final temperature - initial temperature) = (59.4 °C - 62.9 °C)
Substituting the given values, we can calculate the heat gained by the water.
Q_water = m_water * 4.18 J/g°C * (59.4 °C - 62.9 °C)
Since the heat lost by the iron rod is equal to the heat gained by the water, we can set Q_iron equal to Q_water and solve for m_water.
32.3 g * 0.45 J/g°C * (59.4 °C - 22.4 °C) = m_water * 4.18 J/g°C * (59.4 °C - 62.9 °C)
Simplifying the equation and solving for m_water will give you the unknown mass of water in this problem.
I suppose you want to know the mass of H2O but you don't ask it directly.
Heat lost by Fe + heat gained by H2O = 0
Note: sph = specific heat
[mass Fe x sph Fe x (Tfinal-Tinitial)] + [mass H2O x sph H2O x (Tfinal-Tinitial)] = 0