A 223.0 g piece of lead is heated to 85.0oC and then dropped into a calorimeter containing 577.0 g of water that initally is at 18.0oC. Neglecting the heat capacity of the container, find the final equilibrium temperature (in oC) of the lead and water.
Please I need the answer and the equation used to find the answer thanks!
To find the final equilibrium temperature of the lead and water mixture, we can use the principle of conservation of energy. The heat lost by the lead will be equal to the heat gained by the water in the calorimeter.
The heat lost by the lead can be calculated using the formula:
Q_lead = m_lead * c_lead * ΔT_lead,
where Q_lead is the heat lost by the lead, m_lead is the mass of the lead (in grams), c_lead is the specific heat capacity of lead (which is 0.13 J/g°C), and ΔT_lead is the change in temperature of the lead in Celsius.
The heat gained by the water can be calculated 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 mass of the water (in grams), c_water is the specific heat capacity of water (which is 4.18 J/g°C), and ΔT_water is the change in temperature of the water in Celsius.
Since the heat lost by the lead is equal to the heat gained by the water, we can set these two equations equal to each other:
m_lead * c_lead * ΔT_lead = m_water * c_water * ΔT_water.
Now, let's plug in the given values:
m_lead = 223.0 g
c_lead = 0.13 J/g°C
ΔT_lead = (final temperature - 85.0°C) (We'll solve for this later)
m_water = 577.0 g
c_water = 4.18 J/g°C
ΔT_water = (final temperature - 18.0°C)
Now we will rearrange the equation to solve for the final temperature (ΔT_lead):
m_lead * c_lead * ΔT_lead = m_water * c_water * ΔT_water
(223.0 g) * (0.13 J/g°C) * ΔT_lead = (577.0 g) * (4.18 J/g°C) * (final temperature - 18.0°C)
Simplifying the equation:
(29.0 J/°C) * ΔT_lead = (2404.26 J/°C) * (final temperature - 18.0°C)
Dividing both sides of the equation by (29.0 J/°C):
ΔT_lead = (2404.26 J/°C) * (final temperature - 18.0°C) / (29.0 J/°C)
Now we can solve for the final temperature by plugging in the values and rearranging the equation:
(223.0 g) * (0.13 J/g°C) * ΔT_lead = (577.0 g) * (4.18 J/g°C) * (final temperature - 18.0°C)
Now solve for ΔT_lead:
ΔT_lead = [(577.0 g) * (4.18 J/g°C) * (final temperature - 18.0°C)] / [(223.0 g) * (0.13 J/g°C)]
Finally, plug in the calculated value of ΔT_lead into the equation:
ΔT_water = (final temperature - 18.0°C)
and solve for the final temperature.
This equation relates the change in temperature of the lead (ΔT_lead) to the final temperature. By substituting this value into the equation for the change in temperature of the water (ΔT_water), we can solve for the final equilibrium temperature of the lead and water mixture.