When one mole of C6H6 is burned, 3.27 MJ of heat is produced. When the heat from burning 7.59 g of C6H6 is added to 5.69 kg of water at 21.0°C, what is the final temperature of the water?
q=m*c*deltaT
q=heat produced x (7.59)/(molar mass C6H6)
m=mass of water
c=specific heat of water=4.179J/gC
deltaT = T_final - T_initial
T_initial = 20 degreeC
20
To find the final temperature of the water, we need to use the equation:
q = m * c * ΔT
where:
q = heat energy gained or lost by the water (in joules)
m = mass of water (in kg)
c = specific heat capacity of water (approximately 4.18 J/g°C or 4.18 kJ/kg°C)
ΔT = change in temperature of water (final temperature - initial temperature)
First, let's convert the given masses into kilograms:
Mass of water = 5.69 kg
Mass of C6H6 = 7.59 g = 0.00759 kg
Next, we need to calculate the heat released by burning 7.59 g of C6H6:
Heat released = 3.27 MJ = 3.27 × 10^6 J
Now, we can calculate ΔT for the water using the equation:
q = m * c * ΔT
Rearranging the equation, we have:
ΔT = q / (m * c)
ΔT = (3.27 × 10^6 J) / (5.69 kg * 4.18 kJ/kg°C)
After performing the calculations, we find that ΔT ≈ 154.87°C.
Finally, we can determine the final temperature by adding ΔT to the initial temperature of 21.0°C:
Final temperature = 21.0°C + 154.87°C
Hence, the final temperature of the water will be approximately 175.87°C.