The following equation is the balanced combustion reaction for C6H6:
2C6H6(L) + 15O2(g) --> 12CO2(g)+6H2O(L)+6543 kj
If 7.700 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temperature of the water?
basically you want use a formula...then throw your chem book out the window and run.
To find the final temperature of the water, we can use the concept of heat transfer and the equation:
q = m * c * ΔT
where:
q is the heat transfer (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
First, let's calculate the amount of heat produced from burning 7.700 g of C6H6 using the given enthalpy change (ΔH) value:
ΔH = 6543 kJ
We know that 1 kJ is equal to 1000 J. Therefore, the heat produced can be calculated as:
heat = ΔH * (7.700 g C6H6 / molar mass of C6H6)
Next, we need to determine the amount of heat transferred to the water. The specific heat capacity of water, c, is 4.18 J/g°C.
heat transferred to water = m * c * ΔT
Given:
m (mass of water) = 5691 g
initial temperature of water = 21 °C (T1)
final temperature of water (unknown) = T2
The heat transferred to the water can now be expressed as:
heat transferred to water = (5691 g) * (4.18 J/g°C) * (T2 - 21 °C)
Since the heat produced in the combustion reaction is equal to the heat transferred to the water, we can equate the two equations:
ΔH * (7.7 g / molar mass of C6H6) = (5691 g) * (4.18 J/g°C) * (T2 - 21 °C)
Now we can solve for T2 (the final temperature of the water). Rearranging the equation, we have:
T2 - 21 °C = (ΔH * (7.7 g / molar mass of C6H6)) / ((5691 g) * (4.18 J/g°C))
Finally, adding 21 °C to both sides of the equation will give us the final temperature of the water, T2.