DrBob222, I got an answer of 33 and 24.23 as the temperature but it is incorrect. I looked at your response to David and unfortunately it did not help me. Would you please take a look at this question and help me out. Thanks.
If the heat from burning 5.800 g of C6H6 is added to 5691 g of water at 21 °C, what is the final temperature of the water?
2C6H6(l) +15O2 (g)----> 12CO2(g)+6H2O(l) +6542
heat from burning 5.8 g benzene is
6542 x (5.800/2*78.114) = 242,873.2 J.
242,873.2 = 5691 g x 4.184 J/g x (Tf-21)
242,873.2 = 23,811.14T - 500,034.02
T = (500,034.02 + 242,873.2)/23,811.14
T = about 31.19999 which I would round to 31.20 (which is consistent with my estimate with David of 33. I did make a typo on his post which I hope he caught. That is not 293,000, it is closer to 243,000. I hit a 9 instead of a 4. I assume the 21 C is 21.00. There are 4 s.f. everywhere else in the problem, including the 5.800 so 31.20 should do it. You should go through this carefully and confirm all of the numbers. The method is ok but I could have made a calculator error. Check carefully for significant figure issues. I not that many students on this board have the right method and numbers BUT they fail to type in the right number of s.f. Those data bases are so unforgiving. Hope this helps (and works).
Thank you! It did and i found my error a minute before you posted this. Once again, I always appreciate your help DrBob222 :)
To find the final temperature of the water, you need to use the equation for heat transfer:
q = m•c•ΔT
Where:
q = heat transferred
m = mass
c = specific heat capacity
ΔT = change in temperature
In this case, the q value represents the heat released by burning the C6H6 and transferred to the water. The mass of the water is given as 5691 g, and the initial temperature is 21 °C. The final temperature is the unknown we need to solve for.
To proceed, we need to calculate the heat released by burning the C6H6. The balanced chemical equation shows that 2 moles of C6H6 release heat. We can start by converting the mass of C6H6 to moles. The molar mass of C6H6 is 78.11 g/mol, so the number of moles is:
moles C6H6 = mass C6H6 / molar mass C6H6
= 5.800 g / 78.11 g/mol
≈ 0.07424 mol
Since 2 moles of C6H6 release q amount of heat, we can set up a proportion to find the heat released:
q released / 2 moles = x g released / 0.07424 moles
Rearranging the equation, we have:
q released = (2 moles • x g released) / 0.07424 moles
To find x, we need to know the value of q released. Unfortunately, the value 6542 at the end of the equation provided does not seem to represent any unit or value related to heat. Can you provide the correct value for q released?
Once we have the value of q released, we can substitute it into the heat transfer equation along with the mass of water and solve for the change in temperature (ΔT). Adding this change to the initial temperature will give us the final temperature of the water.