A mass 20.0g of petrol was burnt in air. The heat produced was used to heat 2.5 litres of water. Given that the heat value of petrol is 43640 KJ/Kg, what was the temperature change in water?
oops. That's 43640 J/MOL of the reaction and the rxn is for two moles so heat produced will be twice what I had.
2*43640 J/g x 20 g/2(114) = ? and I get approximately 7600 J. Use that new upgraded number you get for q = mass H2O x specific heat H2O x delta T. Sorry about that.
Well, isn't that quite the hot topic! Let's see if we can boil it down for you. To find the temperature change in the water, we need to use the equation Q = mcΔT, where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's convert the mass of water from liters to grams. Since 1 liter of water is equivalent to 1000 grams, 2.5 liters would be 2500 grams.
Now, we know that the specific heat capacity of water is approximately 4.18 J/g°C. However, since the heat value of petrol is given in KJ/Kg, we need to convert KJ to J and Kg to grams.
Since 1 KJ is equivalent to 1000 J, and we have 20 grams of petrol, the total heat energy produced by the petrol is 20 grams × 43640 KJ/Kg × 1000 J/KJ = 872,800 J.
Using the equation Q = mcΔT, we can plug in the known values:
872,800 J = 2500 grams × 4.18 J/g°C × ΔT
Solving for ΔT:
ΔT = 872,800 J / (2500 grams × 4.18 J/g°C)
ΔT ≈ 83.1°C
So, the temperature change in the water is approximately 83.1°C. That's a steamy result!
I want to know how can solution a temperature of change in water
To find the temperature change in water, we can use the formula:
Q = mcΔT
where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, we need to convert the mass of water from liters to grams. Since 1 liter of water has a mass of 1000 grams, 2.5 liters of water would have a mass of 2500 grams.
Next, we can calculate the amount of heat transferred using the formula:
Q = heat value of petrol * mass of petrol
Since the heat value of petrol is given as 43640 KJ/Kg and the mass of petrol is 20.0 g:
Q = 43640 KJ/Kg * 20.0 g
Now we can substitute the values into the formula to find the temperature change:
43640 KJ/Kg * 20.0 g = 872,800 KJ
Q = mcΔT
872,800 KJ = 2500 g * c * ΔT
We know that specific heat capacity of water (c) is approximately 4.18 J/g°C.
Converting KJ to J:
872,800 KJ * 1000 J/KJ = 872,800,000 J
Now we can substitute the values into the formula and solve for ΔT:
872,800,000 J = 2500 g * 4.18 J/g°C * ΔT
Dividing both sides of the equation by (2500 g * 4.18 J/g°C):
ΔT = (872,800,000 J) / (2500 g * 4.18 J/g°C)
Calculating the result:
ΔT = 83.28 °C
Therefore, the temperature change in the water is approximately 83.28 °C.
The major problem with the problem is that petrol is not a pure substance; i.e., it is a mixture of several hydocarbons. Most problems of this type ASSUME that we are dealing with octane; i.e., C8H18 and I will assume that too in order to work the problem.
2C8H18 + 25O2 --> 16CO2 + 18H2O
molar mass C8H18 = 114
The problem gives 43,460 J/g of heat produced, therefore, 20 g of C8H18 will produce how much heat? That will be 43640 J/g x 20 g/2(114) = ? and I gt approximately 3800 J but you should use a more accurate value. How much will that change the temperature of the 2.5 L H2O. We know the specific heat H2O is 4.18 J/g*C so
q = mass x specific heat H2O x delta T
Substitute 3800 (use your more accurate number) = 2,500 g x 4.18 J/g*C x delta T and solve for delta T.
Post your work if you get stuck.