Six grams of water changes from liquid to vapor at a pressure of one atmosphere. In the process, the volume changes from 1.3 cm3 to 2393 cm3. The latent heat of vaporization of water is 2.26 × 106 J/kg.
Find the work done.
Answer in units of J
Find the increase in internal energy of the
water.
Answer in units of J
would appreciate help
To find the work done during the process, we can use the formula:
Work = Pressure × Change in Volume
Given:
Initial volume (Vi) = 1.3 cm^3
Final volume (Vf) = 2393 cm^3
Pressure (P) = 1 atmosphere
First, we need to convert the volumes from cm^3 to m^3 since the SI unit for volume is cubic meters.
Vi = 1.3 cm^3 = 1.3 × 10^-6 m^3
Vf = 2393 cm^3 = 2393 × 10^-6 m^3
Now we can calculate the change in volume:
Change in Volume = Vf - Vi
Substituting the values:
Change in Volume = 2393 × 10^-6 m^3 - 1.3 × 10^-6 m^3
Next, we substitute the values into the formula for work:
Work = P × Change in Volume
Work = 1 atmosphere × (2393 × 10^-6 m^3 - 1.3 × 10^-6 m^3)
To convert atmospheres to pascals (SI unit of pressure), we can use the conversion factor: 1 atmosphere = 101325 pascals.
Now we can calculate the work:
Work = 101325 pascals × (2393 × 10^-6 m^3 - 1.3 × 10^-6 m^3)
Once you perform the calculation, you will obtain the work done in pascal-cubic meters (Pa·m^3). You can convert it to joules (J) since 1 Pa·m^3 is equivalent to 1 J.
To find the increase in internal energy of the water, we can use the formula:
Change in Internal Energy = Mass × Latent Heat of Vaporization
Given:
Mass (m) = 6 grams = 6 × 10^-3 kg
Latent Heat of Vaporization (L) = 2.26 × 10^6 J/kg
Now we can calculate the change in internal energy:
Change in Internal Energy = 6 × 10^-3 kg × 2.26 × 10^6 J/kg
Performing the calculation will give you the increase in internal energy of the water in joules (J).