A 14.9g of lead at 92.5C is dropped into a calorimeter with 165g of water at 20.0C. The final temperature of the system is 20.2C. What is the lead's specific heat?
-Could I get help setting this up? I'm not sure how I'm supposed to set it up.
heat lost by Pb + heat gained by water = 0
[mass Pb x specific heat Pb x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
The only unknown is sp. h. Pb.
To solve this problem, we can apply the principle of heat transfer, which states that the heat lost by the lead will be equal to the heat gained by the water in the calorimeter. This can be expressed mathematically as:
Q_lost = Q_gained
The heat lost by the lead (Q_lost) can be calculated using the formula:
Q_lost = m_lead * c_lead * ΔT_lead
Where:
m_lead is the mass of the lead
c_lead is the specific heat of lead
ΔT_lead is the change in temperature of the lead (final temperature - initial temperature of the lead)
The heat gained by the water (Q_gained) can be calculated using the formula:
Q_gained = m_water * c_water * ΔT_water
Where:
m_water is the mass of the water
c_water is the specific heat of water
ΔT_water is the change in temperature of the water (final temperature - initial temperature of the water)
Since the final temperature (20.2°C) of the system is the same as the initial temperature of the water, ΔT_water will be 0. Therefore, the equation becomes:
Q_lost = Q_gained
m_lead * c_lead * ΔT_lead = m_water * c_water * 0
From here, we can solve for c_lead:
c_lead = (m_water * c_water * 0) / (m_lead * ΔT_lead)
Substituting the given values into the equation, we have:
c_lead = (165g * 4.18J/g°C * 0) / (14.9g * (20.2°C - 92.5°C))
c_lead = 0
Therefore, the specific heat of lead is 0.