A mass has 20.0g of petrol was burnt in air the heat produce wlalsl usedto heat 2.5litter of water given that
the heat value of petrol is 43640kjkg what was the temperature change in water?specific heat capàcity of waters.4.18kjkg-1k-1density of water 1000kg3
Two things you should be aware of. First, it is 2.5 LITERS (not litter).
Second, the density of water is 1000 kg/m3.
First, how much heat do we get from burning 20.0 g petrol. That is
43,640 kJ x (20.0 g/1000 g) = 872.8 kJ. How will the temperature of the water change. That is q = mass H2O x specific heat H2O x delta T
872.8 kJ = 2.5 kg x 4.18 kJ/kg*C x delta T.
Solve for delta T.
Post your work if you get stuck.
Well, well, well, looks like we have a combustion situation here! Let's see what we can do with some clowny calculations.
First things first, let's convert the mass of petrol from grams to kilograms. 20.0g of petrol is equal to 0.020kg of petrol. Easy peasy.
Now, this 0.020kg of petrol, with its heat value of 43640kJ/kg, was burned. And hey, we know that the heat produced was used to heat 2.5 liters of water. But hold up, we need to convert those liters to kilograms because the heat value is given per kilogram.
To convert liters to kilograms, we need to use the density of water. So, knowing that the density of water is 1000kg/m^3 (or 1kg/L), we can say that 2.5 liters of water weigh 2.5kg.
Alright, now that we have all the necessary information, let's calculate the temperature change in the water using the specific heat capacity formula:
Heat Produced = Mass of Water x Specific Heat Capacity x Temperature Change
Substituting the values, we get:
43640kJ/kg x 0.020kg = 2.5kg x 4.18kJ/kg/K x Temperature Change
Now we just solve for the temperature change:
Temperature Change = (43640kJ/kg x 0.020kg) / (2.5kg x 4.18kJ/kg/K)
*clown math mode activated*
Temperature Change = (872.8kJ) / (10.45kJ/K)
Temperature Change = 83.55673758865247 K
*clown math mode deactivated*
So, according to my funny calculations, the temperature change in the water is approximately 83.56 Kelvin. I hope this clowny answer brightened your day a bit!
To calculate the temperature change in water, we need to use the equation:
Q = m * c * ΔT
Where:
Q = heat energy (in joules)
m = mass of water (in kilograms)
c = specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT = temperature change in water (in degrees Celsius)
First, let's convert the mass of water from liters to kilograms:
2.5 liters of water = 2.5 kg (since the density of water is 1000 kg/m³)
Next, we need to calculate the heat energy produced by burning the petrol:
Heat energy = mass of petrol * heat value of petrol
Heat energy = 0.02 kg * 43640 kJ/kg = 872.8 kJ
Now, we can rearrange the equation to calculate the temperature change:
ΔT = Q / (m * c)
ΔT = 872.8 kJ / (2.5 kg * 4.18 kJ/(kg*C)) = 83.5 °C
Therefore, the temperature change in the water is 83.5 degrees Celsius.
To find the temperature change in 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
ΔT is the change in temperature
First, let's convert the volume of water from liters to kilograms.
Given:
Volume of water = 2.5 liters
Density of water = 1000 kg/m^3
Density = mass/volume
Mass = density * volume
Mass of water = 1000 kg/m^3 * 2.5 liters = 1000 kg/m^3 * 0.0025 m^3 = 2.5 kg
Now, let's calculate the heat energy released by burning the petrol.
Given:
Mass of petrol burned = 20.0 g
Heat value of petrol = 43640 kJ/kg
Energy released = mass * heat value
Energy released = 20.0 g * 43640 kJ/kg = 872,800 J
Now we have all the necessary values to find the temperature change in water.
Q = mcΔT
Rearranging the formula, we get:
ΔT = Q / (mc)
Plugging in the values:
ΔT = 872,800 J / (2.5 kg * 4.18 kJ/kg-1K-1)
ΔT = 872,800 J / (10.45 kJ/K)
ΔT ≈ 83.6 K
Therefore, the temperature change in the water is approximately 83.6 Kelvin.