A horizontal piston contains 34kg of water that is saturated at 36ºC. Heat is transferred to the water until 80% of the water vaporizes If the piston moves very slowly, calculate the work done by
surroundings on the system during
this process.
To calculate the work done by the surroundings on the system during this process, we need to know the change in volume of the system.
First, let's find the initial volume of the system. We know that the water is saturated at 36ºC. At this temperature, we can look up the specific volume of saturated water vapor in a steam table. Let's assume the initial specific volume is v1.
Next, we need to find the final volume of the system after 80% of the water has vaporized. To do this, we can use the ideal gas law. Since 80% of the water has vaporized, we can assume it has turned into steam and behaves like an ideal gas. The mass of the remaining water is 34 kg - (0.8 * 34 kg) = 6.8 kg. We can assume the volume of the vaporized water is negligible compared to the remaining water.
Using the ideal gas law, PV = mRT, where P is the pressure, V is the volume, m is the mass, R is the gas constant, and T is the temperature, we can solve for the final volume V2.
Assuming the pressure and temperature remain constant during this process, we can rewrite the equation as V1 = (m1 * R * T) / P and V2 = (m2 * R * T) / P, where m1 is the initial mass of water, m2 is the final mass of water, R is the gas constant, and T is the temperature.
Since the piston moves very slowly, we can assume the process is quasi-static, and the pressure remains constant. Therefore, the work done by the surroundings on the system is given by the equation:
Work = P * (V2 - V1)
Now we have all the components we need to calculate the work done by the surroundings on the system during this process.