the heat of combustion of propane is -5.47MJ/mol. there is 1.50kg of propane available to burn. how many litres of water could be heated from 18.0 degrees Celcius to 63.0 degrees Celcius using the heat from the burning propane. assume that all heat lost by the propane reaction is trapped by the water. you will need the specific heat capacity and density of water.
To find the amount of water that can be heated using the heat from burning propane, we need to calculate the amount of heat released by the combustion and then use that to determine how much the temperature of the water will increase.
First, let's convert the mass of propane from kilograms to moles. The molar mass of propane (C3H8) is 44.1 g/mol, so the number of moles can be calculated as follows:
1.50 kg = 1500 g
Number of moles = 1500 g / 44.1 g/mol = 34.01 mol
Next, we will calculate the energy released during the combustion using the heat of combustion and the number of moles of propane:
Energy released = Heat of combustion × Number of moles
Energy released = -5.47 MJ/mol × 34.01 mol = -185.8687 MJ
Now, we will calculate the amount of heat required to raise the temperature of the water from 18.0 °C to 63.0 °C. The specific heat capacity of water is approximately 4.18 J/g°C, and the density of water is approximately 1 g/mL or 1000 kg/m³:
Mass of water = Density × Volume
Volume of water = Mass of water / Density
Volume of water = (Mass of water / Density of water) = (Mass of water / 1000)
Using the specific heat capacity, we can calculate the amount of heat required:
Heat required = Mass of water × Specific heat capacity × Temperature change
To find the mass of water:
Mass of water = Volume of water × Density of water
Mass of water = (Volume of water × 1000)
Let's calculate the volume of water required first:
Mass of water = (Volume of water × 1000)
Mass of water = (Volume of water × 1000)
Volume of water = Mass of water / 1000
Now, we can calculate the heat required:
Heat required = Mass of water × Specific heat capacity × Temperature change
Heat required = (Mass of water / 1000) × 4.18 J/g°C × (63.0 °C - 18.0 °C)
Finally, we can calculate the volume of water that can be heated using the heat released by the combustion:
Energy released = Heat required
-185.8687 MJ = (Mass of water / 1000) × 4.18 J/g°C × (63.0 °C - 18.0 °C)
Simplifying the equation gives:
-185.8687 × 10^6 J = (Mass of water / 1000) × 4.18 × (63.0 - 18.0)
Solving for Mass of water:
Mass of water = (-185.8687 × 10^6) / (4.18 × 45)
Let's calculate that value:
Mass of water ≈ -9.25 × 10^6 g
Since negative mass is not physically meaningful, it seems there may be an error in the calculations. Let's review the steps and parameters provided to identify any errors.
To solve this problem, we need to calculate the amount of heat energy released by the combustion of propane and then use that value to determine the amount of water that can be heated.
Step 1: Calculate the heat energy released by the combustion of propane.
The heat of combustion of propane is given as -5.47 MJ/mol. First, we need to convert the mass of propane to moles.
Given:
Mass of propane = 1.50 kg
Molar mass of propane (C3H8) = 44.1 g/mol
Converting kg to g:
1.50 kg * 1000 g/kg = 1500 g
Converting g of propane to moles:
1500 g / 44.1 g/mol = 34.01 mol (rounded to two decimal places)
To calculate the heat energy released, we multiply the number of moles by the heat of combustion:
34.01 mol * -5.47 MJ/mol = -185.79 MJ (rounded to two decimal places)
Note that the negative sign indicates that heat is released.
Step 2: Calculate the amount of water that can be heated.
Given:
Initial temperature of water (Ti) = 18.0 degrees Celsius
Final temperature of water (Tf) = 63.0 degrees Celsius
We need to determine the change in temperature (ΔT) of the water:
ΔT = Tf - Ti
ΔT = 63.0 °C - 18.0 °C = 45.0 °C
Now we need to use the specific heat capacity of water to calculate the amount of heat energy required to raise the temperature.
The specific heat capacity of water is approximately 4.18 J/g°C.
Given:
Density of water = 1.00 g/mL (or 1.00 kg/L)
To calculate the amount of water, we need to convert the mass of water to liters.
Using the density:
1.50 kg / 1.00 kg/L = 1.50 L
Finally, we calculate the amount of heat energy needed:
Heat energy (Q) = mass × specific heat capacity × ΔT
Q = 1.50 L × 1000 g/L × 4.18 J/g°C × 45.0 °C
Q = 282,150 J (rounded to the nearest whole number)
Step 3: Determine the number of liters of water that can be heated.
To determine the number of liters of water that can be heated, we divide the amount of heat energy released by the amount of heat energy required to heat the water:
Number of liters = Heat energy released / Heat energy required
Number of liters = -185.79 MJ / 282,150 J
Number of liters = -658 L (rounded to the nearest whole number)
Since negative volume would not make sense in this context, we take only the magnitude of the volume. Therefore, the answer is that approximately 658 liters of water can be heated from 18.0 °C to 63.0 °C using the heat from the burning propane.