Given that the molar enthalpy of combustion of propane is -2041.9 in an open system, how much water could be heated by 60 degree Celsius by the combustion of 28.5g of propane.
To calculate the amount of water that can be heated by the combustion of propane, we need to determine the amount of heat released by the combustion reaction.
The molar enthalpy of combustion of propane is given as -2041.9 kJ/mol. This means that when one mole of propane is combusted in an open system, -2041.9 kJ of heat is released.
First, we need to calculate the number of moles of propane in 28.5g. The molar mass of propane (C3H8) is 44.1 g/mol. Therefore, the number of moles of propane is given by:
Number of moles = Mass of propane / Molar mass of propane
Number of moles = 28.5g / 44.1 g/mol
Number of moles = 0.647 moles
Now, the amount of heat released by combustion of 0.647 moles of propane can be calculated using the molar enthalpy of combustion:
Heat released = Number of moles * Molar enthalpy of combustion
Heat released = 0.647 mol * -2041.9 kJ/mol
Heat released = -1320.29 kJ
To calculate how much water can be heated by this amount of heat, we will use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q = amount of heat absorbed (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (in °C)
The change in temperature is given as 60°C and we need to find the mass of water (m). Rearranging the equation, we get:
m = Q / (c * ΔT)
Substituting the values into the equation:
m = -1320.29 kJ / (4.184 J/g°C * 60°C)
m = -1320.29 kJ / 250.44 J/g
m ≈ -5.27 g
Since mass cannot be negative, we will take the absolute value:
m ≈ 5.27 g
Therefore, approximately 5.27 grams of water could be heated by 60°C by the combustion of 28.5 grams of propane.
To calculate the amount of water that can be heated by the combustion of propane, we need to determine the heat energy released during the combustion.
First, we need to convert the mass of propane (28.5 g) to moles. The molar mass of propane (C3H8) can be calculated by summing the atomic masses of its constituent elements:
C: 12.01 g/mol
H: 1.01 g/mol
Molar mass of propane (C3H8) = (3 × C) + (8 × H) = (3 × 12.01) + (8 × 1.01) = 44.11 g/mol
Next, we calculate the number of moles of propane:
Number of moles of propane = mass of propane / molar mass of propane
= 28.5 g / 44.11 g/mol
≈ 0.647 mol
The molar enthalpy of combustion of propane represents the heat energy released per mole of propane combusted. In this case, it is given as -2041.9 kJ/mol (negative sign indicates the release of energy).
Now, we can calculate the total heat energy released during the combustion:
Heat energy released = molar enthalpy of combustion × number of moles of propane
= -2041.9 kJ/mol × 0.647 mol
≈ -1321.97 kJ
Since we want to determine the amount of water that can be heated by 60 degrees Celsius, we need to consider the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g·°C.
To calculate the amount of water that can be heated, we can use the equation:
Heat energy released = mass of water × specific heat capacity of water × change in temperature
Rearranging the equation, we can solve for the mass of water:
Mass of water = Heat energy released / (specific heat capacity of water × change in temperature)
Let's assume the change in temperature of the water is 60 degrees Celsius:
Mass of water = -1321970 J / (4.18 J/g·°C × 60 °C)
≈ 5040.05 g
Therefore, the mass of water that can be heated by the combustion of 28.5 g of propane is approximately 5040.05 grams (or about 5.04 kg).
To determine how much water can be heated by the combustion of propane, we need to calculate the amount of heat released during the combustion and then use that information to find the temperature change of the water.
Step 1: Calculate the moles of propane burned.
Given that the molar mass of propane (C3H8) is approximately 44.1 g/mol, we can calculate the number of moles burned:
Number of moles = mass of propane / molar mass of propane
Number of moles = 28.5 g / 44.1 g/mol ≈ 0.646 moles
Step 2: Calculate the heat released during combustion.
The molar enthalpy of combustion of propane is given as -2041.9 kJ/mol. Therefore, the heat released during the combustion of propane can be calculated as follows:
Heat released = molar enthalpy of combustion × number of moles
Heat released = -2041.9 kJ/mol × 0.646 moles ≈ -1318.7 kJ
Step 3: Calculate the heat absorbed by the water.
The amount of heat absorbed by the water is determined by its mass, specific heat capacity, and the temperature change.
Given that the specific heat capacity of water is approximately 4.18 J/g°C, we can use the following equation to calculate the heat absorbed by the water:
Heat absorbed = mass of water × specific heat capacity of water × temperature change
However, we need to convert the heat released from kJ to J and the mass of water from grams to kilograms to maintain consistent units:
Heat absorbed = mass of water (in kg) × specific heat capacity of water × temperature change
Step 4: Rearrange the equation to solve for the mass of water.
Mass of water (in kg) = Heat absorbed / (specific heat capacity of water × temperature change)
Let's assume the temperature change is 60 °C.
Mass of water (in kg) = -1318.7 kJ / (4.18 J/g°C × 60 °C) = -1318.7 kJ / 250.8 J = -5.25 kg
Since mass cannot be negative, it seems that there might be an error in one of the given values or calculations. Please double-check the values and try again.