The molar enthalpy of combustion of ethanol is -1234.8kJ/mol. How much water can be heated from 10.85 degrees celsius to 72.6 degrees celsius by the combustion of 48.2 g of ethanol.
To find out how much water can be heated by the combustion of ethanol, we need to calculate the amount of heat produced by the combustion reaction and then use this value to determine the change in temperature of the water.
1. Calculate the moles of ethanol used:
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Moles of ethanol = mass of ethanol / molar mass of ethanol
Moles of ethanol = 48.2 g / 46.07 g/mol
Moles of ethanol = 1.046 mol
2. Calculate the heat produced by the combustion of ethanol:
Heat produced = Moles of ethanol * Molar enthalpy of combustion
Heat produced = 1.046 mol * (-1234.8 kJ/mol)
Heat produced = -1290.08 kJ
3. Calculate the amount of heat transferred to the water:
Q = mcΔT
where Q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C (or 4.18 kJ/kg°C).
Mass of water = molar mass of water * moles of water
molar mass of water = 18.015 g/mol
Mass of water = 18.015 g/mol * Moles of water
Since we don't know the moles of water yet, we have to assume that all the heat produced by the combustion is transferred to the water.
4. Calculate the moles of water using the equation for molar enthalpy of combustion:
Molar enthalpy of combustion of ethanol = heat produced / moles of water
-1290.08 kJ = heat produced / moles of water
moles of water = heat produced / molar enthalpy of combustion of ethanol
moles of water = -1290.08 kJ / -1234.8 kJ/mol
moles of water ≈ 1.045 mol
5. Calculate the mass of water:
Mass of water = moles of water * molar mass of water
Mass of water = 1.045 mol * 18.015 g/mol
Mass of water ≈ 18.8 g
6. Finally, calculate the change in temperature of the water:
Q = mcΔT
Q = -1290.08 kJ
m = 18.8 g
c = 4.18 kJ/kg°C
ΔT = ? (change in temperature)
-1290.08 kJ = (18.8 g * 4.18 kJ/kg°C) * ΔT
ΔT = -1290.08 kJ / (18.8 g * 4.18 kJ/kg°C)
ΔT ≈ -18.6°C
Since the change in temperature is negative, it means that the water will cool down during the combustion of ethanol.
Therefore, the water can be cooled from 10.85°C to approximately -7.75°C by the combustion of 48.2 g of ethanol.
To calculate the amount of water that can be heated by the combustion of ethanol, we need to use the principle of conservation of energy.
First, we need to determine the amount of heat released by the combustion of 48.2 g of ethanol. This can be calculated using the molar enthalpy of combustion.
1. Convert the mass of ethanol to moles:
To convert the mass of ethanol (48.2 g) to moles, we need to know its molar mass. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol. Using this information, we can calculate the number of moles of ethanol:
moles of ethanol = mass of ethanol / molar mass of ethanol
= 48.2 g / 46.07 g/mol
≈ 1.046 mol
2. Calculate the heat released by the combustion of ethanol:
The molar enthalpy of combustion of ethanol (-1234.8 kJ/mol) represents the heat released when 1 mole of ethanol is completely burned. Therefore, we can calculate the heat released by the combustion of 48.2 g of ethanol as follows:
heat released = molar enthalpy of combustion * moles of ethanol
= -1234.8 kJ/mol * 1.046 mol
≈ -1290.1 kJ
Since the negative sign indicates that heat is released, we take the absolute value and consider it as the quantity of heat released.
3. Calculate the heat absorbed by the water:
To calculate the heat absorbed by the water, we can use the formula:
q = m * c * ΔT
where:
- q is the heat absorbed (in joules)
- m is the mass of the water (in grams)
- c is the specific heat capacity of water (4.184 J/g·°C)
- ΔT is the change in temperature (in °C)
Let's plug in the values into the equation:
q = m * c * ΔT
We know:
- Initial temperature (Ti) = 10.85 °C
- Final temperature (Tf) = 72.6 °C
- Change in temperature (ΔT) = Tf - Ti
Assuming the specific heat capacity of water remains constant over this temperature range, we can calculate the amount of water (m) using the equation:
m = q / (c * ΔT)
Let's substitute the known values into the equation:
m = (-1290.1 kJ) / (4.184 J/g·°C * (72.6 - 10.85) °C)
Now we can convert the units to grams before calculating the final value:
m = (-1290.1 kJ * 1000 g/kg) / (4.184 J/g·°C * 61.75 °C)
Simplifying the equation gives us:
m ≈ 32.2 g
Therefore, approximately 32.2 grams of water can be heated from 10.85 degrees Celsius to 72.6 degrees Celsius by the combustion of 48.2 grams of ethanol.
To determine how much water can be heated, we need to calculate the amount of heat released by the combustion of 48.2 g of ethanol using its molar enthalpy of combustion. Then, we can use this heat value to calculate the temperature increase of the water.
Step 1: Calculate the number of moles of ethanol used.
The molar mass of ethanol (C2H5OH) is 46.07 g/mol. We can use this value to calculate the number of moles of ethanol:
Number of moles = Mass of ethanol / Molar mass of ethanol
= 48.2 g / 46.07 g/mol
≈ 1.046 mol (rounded to 3 decimal places)
Step 2: Calculate the heat released by the combustion of ethanol.
The molar enthalpy of combustion of ethanol is given as -1234.8 kJ/mol. We can use this value to calculate the heat released:
Heat released = Molar enthalpy of combustion * Number of moles
= -1234.8 kJ/mol * 1.046 mol
≈ -1290.0768 kJ (rounded to 4 decimal places)
Step 3: Calculate the amount of water that can be heated.
The specific heat capacity of water is approximately 4.18 J/g·°C (rounded to 2 decimal places).
To calculate the amount of heat required to heat the water, we can use the formula:
Heat required = Mass of water * Specific heat capacity * Temperature increase
Rearranging the formula, we get:
Mass of water = Heat required / (Specific heat capacity * Temperature increase)
The initial temperature (Ti) of the water is 10.85 °C, and the final temperature (Tf) is 72.6 °C. The temperature increase is Tf - Ti.
Temperature increase = Tf - Ti
= 72.6 °C - 10.85 °C
= 61.75 °C (rounded to 2 decimal places)
Heat required = -1290.0768 kJ (converted to J)
= -1290.0768 kJ * 1000 J/1 kJ
= -1290076.8 J (rounded to 1 decimal place)
Mass of water = (Heat required) / (Specific heat capacity * Temperature increase)
= -1290076.8 J / (4.18 J/g·°C * 61.75 °C)
≈ -517.4215 g (rounded to 4 decimal places)
The negative sign indicates that the water was heated by the combustion of ethanol. However, it is not possible for the mass of water to have a negative value, so we take the absolute value:
Mass of water = |-517.4215 g|
≈ 517.4215 g (rounded to 4 decimal places)
Therefore, approximately 517.4215 grams of water can be heated from 10.85 degrees Celsius to 72.6 degrees Celsius by the combustion of 48.2 g of ethanol.