When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 5.35 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?
How much heat is released? That's
3.27E6 J x (5.35/molar mass C6H6) = ?
Then q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
Convert kg H2O to grams. Use specific heat in J/g*C
To solve this problem, we will use the concept of heat transfer and heat capacity.
1. First, let's calculate the heat transferred to the water by using the formula:
Q = mcΔT
Q: heat transferred to the water (in joules)
m: mass of the water (in kilograms)
c: specific heat capacity of water (4.18 J/g°C or 4180 J/kg°C)
ΔT: change in temperature (final temperature - initial temperature)
Since we have the mass of water in grams, we need to convert it to kilograms:
m = 5.69 kg × 1000 = 5690 g
Now we can calculate the heat transferred to the water:
Q = (5.69 kg) × (4180 J/kg°C) × ΔT
2. Next, we need to calculate the heat released by burning 5.35 g of C6H6. We know that when one mole of C6H6 is burned, 3.27 MJ of heat is produced. To find the heat released by 5.35 g of C6H6, we can use the molar mass of C6H6 which is approximately 78 g/mol.
mol C6H6 = 5.35 g ÷ 78 g/mol
Then, we can calculate the heat released:
Heat released = (mol C6H6) × (3.27 MJ/mol) × (1 MJ/10^6 J)
3. Now we have the heat transferred to the water and the heat released by burning C6H6. According to the law of energy conservation, these two values should be equal:
(5.69 kg) × (4180 J/kg°C) × ΔT = Heat released
We can rearrange this equation to solve for ΔT:
ΔT = Heat released / [(5.69 kg) × (4180 J/kg°C)]
4. Calculate the value of ΔT:
ΔT = (Heat released) / [(5.69 kg) × (4180 J/kg°C)]
5. Finally, we can find the final temperature of the water by adding ΔT to the initial temperature:
Final temperature = Initial temperature + ΔT
Given that the initial temperature is 21.0°C, we can substitute the value of ΔT to find the final temperature of the water.