After adding a hot piece of titanium to a cup containing 300.0 mL of 30.0 °C water, the final temperature of the solution is 41.2 °C and the final volume of the solution is 400.0 mL. The density of titanium is 4.50 g/cm3. If no energy is lost to the surroundings, what was the initial temperature of the piece of titanium? (The mass of the water is 300.0g)
heat lost by T + heat gained by water = 0
[mass Ti x specific heat Ti x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
You have all of the numbers to substitute except for mass Ti. You solve for that from the density. The volume is 100 cc (400 final volume H2O-300 initial volume H2O = 100 mL = 100 cc Ti metal). Use density of mass/volume. You have volume and density, solve for mass.
To find the initial temperature of the piece of titanium, we can use the principle of conservation of energy and heat transfer equations.
First, let's calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
where:
q_water is the heat gained by the water
m_water is the mass of the water (300.0 g)
c_water is the specific heat capacity of water (4.184 J/g°C)
ΔT_water is the change in temperature of the water (final temperature - initial temperature)
Next, let's calculate the heat lost by the titanium:
q_titanium = m_titanium * c_titanium * ΔT_titanium
where:
q_titanium is the heat lost by the titanium
m_titanium is the mass of the titanium
c_titanium is the specific heat capacity of titanium (0.523 J/g°C, assuming room temperature)
ΔT_titanium is the change in temperature of the titanium (final temperature - initial temperature)
Since no energy is lost to the surroundings, the heat gained by the water is equal to the heat lost by the titanium:
q_water = q_titanium
Setting up the equation:
m_water * c_water * ΔT_water = m_titanium * c_titanium * ΔT_titanium
We have the following values:
- m_water = 300.0 g
- c_water = 4.184 J/g°C
- ΔT_water = 41.2 °C - 30.0 °C
- ΔT_titanium = 41.2 °C - T_initial (where T_initial is the initial temperature of the titanium)
We are also given the density of titanium: density_titanium = 4.50 g/cm3. We can use this to calculate the mass of the titanium:
m_titanium = (volume_titanium) * (density_titanium)
Since the water and titanium are mixed, the final volume of the solution is 400.0 mL. Therefore, we can calculate the volume of the titanium:
volume_titanium = final volume - volume_water
volume_titanium = 400.0 mL - 300.0 mL
Now we can substitute the values into the equation and solve for T_initial:
m_water * c_water * ΔT_water = m_titanium * c_titanium * ΔT_titanium
(300.0 g) * (4.184 J/g°C) * (41.2 °C - 30.0 °C) = [(400.0 mL - 300.0 mL) * (4.50 g/cm3)] * (0.523 J/g°C) * (41.2 °C - T_initial)
Simplify and solve for T_initial.