1)Suppose a piece of lead with a mass of 14.9 g at a temperature of 92.5 degrees is dropped into an insulated container of water. The mass of water is 165 g and its temperature before adding the lead is 20.0 degrees. What is the final temperature of the system?
2)What are the proportions by mass of the elements in the compound sodium sulfate, Na2SO4?
1.
heat lost piece of lead + heat gained by wter = 0
heat lost by lead = mass x specific heat lead x (Tf-Ti) where Tf is final T (in this case the unknown) and Ti is the initial T of the lead.
heat gained by wter = mass x specific heat water x (Tf - Ti).
Solve for Tf. Post your work if you get stuck.
2.
molar mass Na2SO4 is approximately,
(2 x 23) + 32 + (4 x 16) = 142. You should confirm this by looking up the atomic masses in the periodic table and adding them.
Then
%Na = [(2 x 23)/142]*100
%S = (32/142)*100
%O = [(4 x 16)/142]*100
I have rounded here and there so you need to look these numbers up from the periodic table and recalculate to the number of significant figures your teacher wants.
1) To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the lead will be gained by the water. We can calculate the heat lost by the lead using the formula:
Q_lost = m_lead * c_lead * (T_initial_lead - T_final_sys)
where
m_lead = mass of the lead = 14.9 g
c_lead = specific heat capacity of lead = 0.128 J/g°C (given)
T_initial_lead = initial temperature of the lead = 92.5°C
T_final_sys = final temperature of the system (water + lead)
Next, we can calculate the heat gained by the water using the formula:
Q_gained = m_water * c_water * (T_final_sys - T_initial_water)
where
m_water = mass of water = 165 g
c_water = specific heat capacity of water = 4.184 J/g°C (specific heat capacity of water is always considered as 4.184 J/g°C)
T_initial_water = initial temperature of water = 20.0°C
According to the principle of conservation of energy, Q_lost = Q_gained. Therefore, we can set up the equation:
m_lead * c_lead * (T_initial_lead - T_final_sys) = m_water * c_water * (T_final_sys - T_initial_water)
Plugging in the given values, we can solve for T_final_sys:
14.9 g * 0.128 J/g°C * (92.5°C - T_final_sys) = 165 g * 4.184 J/g°C * (T_final_sys - 20.0°C)
Simplifying the equation:
12.0 g°C * (92.5°C - T_final_sys) = 688.26 g°C * (T_final_sys - 20.0°C)
1104 g°C - 12.0 g°C * T_final_sys = 688.26 g°C * T_final_sys - 13765.2 g°C
Combining like terms:
12.0 g°C * T_final_sys + 688.26 g°C * T_final_sys = 1104 g°C + 13765.2 g°C
700.26 g°C * T_final_sys = 14869.2 g°C
Dividing both sides by 700.26 g°C:
T_final_sys = 21.2°C
Therefore, the final temperature of the system is 21.2°C.
2) To determine the proportions by mass of the elements in sodium sulfate (Na2SO4), we need to calculate the molar masses of each element present.
The molar mass of sodium (Na) = 22.99 g/mol (found on the periodic table)
The molar mass of sulfur (S) = 32.07 g/mol (found on the periodic table)
The molar mass of oxygen (O) = 16.00 g/mol (found on the periodic table)
Now, let's calculate the molar mass of Na2SO4:
Molar mass of Na2SO4 = (2 * Na) + S + (4 * O)
= (2 * 22.99 g/mol) + 32.07 g/mol + (4 * 16.00 g/mol)
= 45.98 g/mol + 32.07 g/mol + 64.00 g/mol
= 142.05 g/mol
To find the proportions by mass of each element, we divide the molar mass of each specific element in Na2SO4 by the molar mass of the entire compound, and then multiply by 100% to get the percentage.
Proportion of sodium (Na):
= (2 * Na molar mass) / Na2SO4 molar mass * 100%
= (2 * 22.99 g/mol) / 142.05 g/mol * 100%
= 32.26%
Proportion of sulfur (S):
= S molar mass / Na2SO4 molar mass * 100%
= 32.07 g/mol / 142.05 g/mol * 100%
= 22.58%
Proportion of oxygen (O):
= (4 * O molar mass) / Na2SO4 molar mass * 100%
= (4 * 16.00 g/mol) / 142.05 g/mol * 100%
= 45.16%
Therefore, the proportions by mass of the elements in sodium sulfate (Na2SO4) are approximately:
- Sodium: 32.26%
- Sulfur: 22.58%
- Oxygen: 45.16%
To solve these questions, we can use the principle of conservation of energy and the concept of heat transfer.
1) To find the final temperature of the system, we need to understand how heat is transferred between the lead and the water. This can be done using the equation:
Q(m1c1ΔT1) = Q(m2c2ΔT2)
where Q represents heat transfer, m represents mass, c represents specific heat capacity, and ΔT represents change in temperature.
In this case, the lead and the water exchange heat until reaching thermal equilibrium, so their total heat transfer is zero:
Q_lead + Q_water = 0
Since Q = mcΔT, we can rearrange the equation above to:
m1c1ΔT1 + m2c2ΔT2 = 0
Plugging in the given values, we get:
(14.9 g)(specific heat capacity of lead)(final temperature - 92.5°C) + (165 g)(specific heat capacity of water)(final temperature - 20.0°C) = 0
Using the specific heat capacity values for lead and water from reference sources, we can solve for the final temperature.
2) To find the proportions by mass of the elements in sodium sulfate (Na2SO4), we need to determine the molar mass of each element and then calculate the mass percentage.
The molar mass of sodium (Na) is approximately 22.99 g/mol, sulfur (S) is approximately 32.07 g/mol, and oxygen (O) is approximately 16.00 g/mol.
The molar mass of the entire compound Na2SO4 is then:
(2 * Na) + S + (4 * O) = 22.99 * 2 + 32.07 + 16.00 * 4 = XX g/mol
To find the mass percentage of each element, we use the following formula:
(molar mass of each element / molar mass of the compound) * 100
For sodium:
(molar mass of sodium / molar mass of the compound) * 100 = (22.99 / XX) * 100 = x%
For sulfur:
(molar mass of sulfur / molar mass of the compound) * 100 = (32.07 / XX) * 100 = y%
For oxygen:
(molar mass of oxygen / molar mass of the compound) * 100 = (16.00 * 4 / XX) * 100 = z%
By substituting the respective molar mass values, you can calculate the mass percentages of each element in sodium sulfate.