Household equipment called clothes vaporizers use water vapor generated by a system of electrical resistances from liquid water. A device with a nominal power of 1,600 Watts was used to iron clothes for 20 minutes, consuming 540 mL of water. In relation to the total energy expenditure of the equipment, the energy expenditure used only to vaporize the water, after it has already reached the boiling temperature, is equivalent to approximately
A. 0.04%. B. 0.062%. C. 4.6%. D. 40%. E. 62%.
We have 540 milli liters of liquid water to become steam. So we have to calculate the energy needed to vaporize that amount of liquid water. But the enthalpy of vaporization is calculated based on moles of the substance. Then we have to convert the volume to Mols. First we’ll have to turn it into grass. We see that the density of the water is 1 gram per milli Liter. So we have 540 times 1, which tells us that the mass is 540 grams. To calculate now the number of moles, we have to divide the mass, by the molar mass, so I have 540 grams, divided by 18 grams per mol, which results in 30 moles.
Now to have the total vaporization energy, we multiply the enthalpy of vaporization to 100 degrees, times the number of moles, resulting in 1200 kilo Joule. So now we have all the energy used to vaporize the water.
But we have the resistance that uses 1600 Watts, a Watt is 1 Joule per second. The time was 20 minutes, each minute has 60 seconds. So the time is 20 minutes times 60 seconds, which is 1200 seconds. To see the total spend, we multiply the Power by the time it was on, which is 1200 seconds, generating 1920 kilo Joules.
The question was how much was spent on vaporization energy, in relation to all applied energy.
For this we divide the total energy of vaporization, by the total energy used. So with this division we have the value of 62 percent of total energy, spent to evaporate the water.
A. 0.04%. B. 0.062%. C. 4.6%. D. 40%. E. 62%.
We have 540 milli liters of liquid water to become steam. So we have to calculate the energy needed to vaporize that amount of liquid water. But the enthalpy of vaporization is calculated based on moles of the substance. Then we have to convert the volume to Mols. First we’ll have to turn it into grass. We see that the density of the water is 1 gram per milli Liter. So we have 540 times 1, which tells us that the mass is 540 grams. To calculate now the number of moles, we have to divide the mass, by the molar mass, so I have 540 grams, divided by 18 grams per mol, which results in 30 moles.
Now to have the total vaporization energy, we multiply the enthalpy of vaporization to 100 degrees, times the number of moles, resulting in 1200 kilo Joule. So now we have all the energy used to vaporize the water.
But we have the resistance that uses 1600 Watts, a Watt is 1 Joule per second. The time was 20 minutes, each minute has 60 seconds. So the time is 20 minutes times 60 seconds, which is 1200 seconds. To see the total spend, we multiply the Power by the time it was on, which is 1200 seconds, generating 1920 kilo Joules.
The question was how much was spent on vaporization energy, in relation to all applied energy.
For this we divide the total energy of vaporization, by the total energy used. So with this division we have the value of 62 percent of total energy, spent to evaporate the water.