q = ,mass H2O x specific heat H2O x delta T
q = 200 g x 4.184 J/g*C x 12 = ? J
q = 200 g x 4.184 J/g*C x 12 = ? J
q = m × c × ∆T
where q represents the energy released, m is the mass of water, c is the specific heat capacity of water, and ∆T is the change in temperature.
Given values:
m (mass of water) = 2.00 x 10^2g
c (specific heat capacity of water) = 4.18 J/g°C (approximate value for liquid water)
∆T (change in temperature) = 37.00°C - 25.00°C = 12.00°C
Now, let's plug in the values:
q = (2.00 x 10^2g) × (4.18 J/g°C) × (12.00°C)
q = 100 × 4.18 × 12
q = 5016 J
Therefore, the system released approximately 5016 Joules of energy.
1. First, we need to calculate the heat transfer for the water. The specific heat capacity of water (c) is 4.18 J/(g°C).
q_water = (m_water) x (c_water) x (ΔT)
= (200.0 g) x (4.18 J/(g°C)) x (37.00°C - 25.00°C)
= 200.0 g x 4.18 J/(g°C) x 12.00°C
= 10056 J
Therefore, the heat transfer for the water is 10056 J.
2. Next, we need to calculate the heat transfer for the octane using the same equation. The molar mass of octane (C8H18) is 114.22 g/mol, so we can calculate the number of moles of octane burned.
n_octane = (m_octane) / (Molar mass_octane)
= 10.0 g / 114.22 g/mol
≈ 0.0874 mol
The enthalpy change (ΔH) for the combustion of octane is -5471 kJ/mol.
q_octane = (n_octane) x (ΔH_octane)
= 0.0874 mol x (-5471 kJ/mol) x (1000 J/1 kJ)
≈ -477 kJ
Therefore, the heat transfer for the octane is -477000 J (negative because energy is released).
3. Finally, we can add the heat transfers for the water and the octane to get the total heat transfer for the system.
q_total = q_water + q_octane
= 10056 J + (-477000 J)
= -466944 J
So, the amount of energy released by the system is approximately -466944 J (negative value indicates energy released).