We have of ice at -11.9°C and it is ultimately completely converted to liquid at 9.9°C. For ice, use a specific heat (Cs) of 2.01 J/gK, for liquid water use a specific heat of 4.18 J/gK, and ΔHfusion = 6.01kJ/mole, Molar mass H20 = 18.01g/mol. If 1 mole of ice at -11.9°C is heated by the combustion of 0.010 mole of CH4(g), what would the final temperature be? ΔHf°(CH4) = -74.6kJ/mol; ΔHf°(CO2)=-393.5kJ/mol; ΔHf°(H2O(g))=-241.8kJ/mol.
Balanced equation:
CH4 + 2O2 = CO2 + 2H2O
dHcomb = (n*dHformation products) - (n*dHformation products)
I ran through this and estimate about 802.5 kJ/mol so the heat emitted by 0.01 mol CH4 will be 8.025 kJ or 8,025 J.
I would do the rest of it in pieces.
It will take
q1 = heat to raise T of solid ice from -11.9 to solid ice at zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = ?
Then it will take 6.01 kJ/mol to melt the ice at zero to liquid H2O at zero. Since you had 1 mol that will take 6.01 kJ or 6010 J and that is q2.
Now you are faced with how much will the remaining heat raise the temperature
so the question is how much heat is remaining. That is
8025 J - q1 - q2 = ? = q3
Finally, plug in q3 and see how much Tf is.
q3 = mass H2O x specific heat H2O x (Tf-Ti). Solve for Tf; you know Ti is zero.
I think the answer is about 20C but that's just a close answer.
What's the Tfinal - Tinitial? i tried (0 - (-11.9)) but I don't think its the right thing.
To find the final temperature, we need to calculate the heat gained by the ice and the heat released by the combustion reaction of CH4.
First, let's calculate the heat gained by the ice using the equation:
q(ice) = m × Cs × ΔT
where:
q(ice) is the heat gained by the ice,
m is the mass of ice,
Cs is the specific heat of ice, and
ΔT is the change in temperature.
Given that we have 1 mole of ice and the molar mass of H2O is 18.01 g/mol, the mass of ice is 18.01 g.
ΔT = final temperature - initial temperature = 9.9°C - (-11.9°C) = 21.8°C
Now, let's calculate the heat gained by the ice:
q(ice) = (18.01 g) × (2.01 J/gK) × (21.8°C)
= 791.5974 J
Next, let's calculate the heat released by the combustion of CH4. We need to consider the balanced equation to determine the number of moles of CH4 consumed and the moles of H2O produced:
CH4 + 2O2 = CO2 + 2H2O
From the balanced equation, we can see that 1 mole of CH4 produces 2 moles of H2O. Therefore, if 0.010 moles of CH4 is burned, it will result in 0.020 moles of H2O.
Now, let's calculate the heat released by the combustion of CH4:
q(CH4) = ΔHf°(CH4) × n(CH4)
where:
q(CH4) is the heat released by the combustion of CH4,
ΔHf°(CH4) is the standard enthalpy of formation of CH4, and
n(CH4) is the number of moles of CH4.
Given ΔHf°(CH4) = -74.6 kJ/mol and n(CH4) = 0.010 moles:
q(CH4) = (-74.6 kJ/mol) × (0.010 moles) × (1000 J/1 kJ)
= -746 J
Since the heat gained by the ice is equal to the heat released by the combustion of CH4, we set up the equation:
q(ice) = q(CH4)
791.5974 J = -746 J
Now we can solve for the final temperature:
q(ice) = m × Cs × ΔT
791.5974 J = (18.01 g) × (4.18 J/gK) × (final temperature - 9.9°C)
Solving for the final temperature:
(final temperature - 9.9°C) = 791.5974 J / ((18.01 g) × (4.18 J/gK))
final temperature - 9.9°C = 11.5584 K
final temperature = 21.4584°C
Therefore, the final temperature would be approximately 21.5°C.