A small “coffee cup” calorimeter contains 110. g of H2O at 22.0°C. A 100. g sample of
lead is heated to 90.0°C and then placed in the water. The contents of the calorimeter come to a
temperature of 23.9°C What is the specific heat of lead?
I don't get which one is Tfinal, and Tinitial for both of them
can you explain it? thank you
[mass Pb x specific heat lead x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0.
To find the specific heat of lead, we can use the formula:
q = m * c * ΔT
where:
q = heat transferred
m = mass of the substance (lead)
c = specific heat of the substance (lead)
ΔT = change in temperature
First, let's find the heat transferred to the water by the lead:
q_water = m_water * c_water * ΔT_water
The mass of water is 110. g, the specific heat of water is 4.18 J/g°C, and the change in temperature is (23.9 - 22.0)°C.
q_water = 110. g * 4.18 J/g°C * (23.9 - 22.0)°C
Next, let's find the heat transferred from the lead to the water:
q_lead = m_lead * c_lead * ΔT_lead
The mass of the lead is 100. g, the specific heat of the lead is what we are trying to find, and the change in temperature is (23.9 - 90.0)°C.
q_lead = 100. g * c_lead * (23.9 - 90.0)°C
Since the total heat transferred to the water is the same as the heat transferred from the lead, we can set them equal to each other:
q_water = q_lead
110. g * 4.18 J/g°C * (23.9 - 22.0)°C = 100. g * c_lead * (23.9 - 90.0)°C
Now, we can solve for the specific heat of lead:
c_lead = (110. g * 4.18 J/g°C * (23.9 - 22.0)°C) / (100. g * (23.9 - 90.0)°C)
c_lead = (110. g * 4.18 J/g°C * 1.9°C) / (100. g * -66.1°C)
c_lead ≈ -0.034 J/g°C
The specific heat of lead is approximately -0.034 J/g°C. Note that the negative sign indicates an inverse relationship between the change in temperature and the heat transferred.
To find the specific heat of lead, we can use the principle of calorimetry. Calorimetry is the measurement of heat transfer between objects. It involves the use of a calorimeter, a device used to measure the heat absorbed or released during a physical or chemical process.
In this problem, the heat gained by the water is equal to the heat lost by the lead:
Heat gained by water = Heat lost by lead
The heat gained or lost can be calculated using the formula:
q = m * c * ΔT
where:
q is the heat gained or lost
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
Let's break down the problem and calculate the heat gained by the water and the heat lost by the lead separately.
1. Heat gained by water:
The mass of water is given as 110. g, and the initial temperature is 22.0°C. The final temperature is 23.9°C, so the change in temperature is:
ΔT_water = (23.9°C - 22.0°C)
Now, we can calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
2. Heat lost by lead:
The mass of lead is given as 100. g, and the initial temperature is 90.0°C. The final temperature is 23.9°C, so the change in temperature is:
ΔT_lead = (23.9°C - 90.0°C)
Now, we can calculate the heat lost by the lead:
q_lead = m_lead * c_lead * ΔT_lead
3. Equating the two equations:
Since the heat gained by the water is equal to the heat lost by the lead, we can set them equal to each other:
q_water = q_lead
m_water * c_water * ΔT_water = m_lead * c_lead * ΔT_lead
You are given the values for mass and temperatures. We need to solve for the specific heat of lead (c_lead), so we rearrange the equation:
c_lead = (m_water * c_water * ΔT_water) / (m_lead * ΔT_lead)
Let's plug in the given values and calculate the specific heat of lead.