Get Help with 2 Questions: Solar Water Heater & Air Pressure | Expert Tips

[ABR124]

Hi

I am stuck on two questions here not sure if this is the right part of the forum to post it in, sorry in advance if it isn't. Ok the questions, my problem is that I'm not sure on the equation to use to solve these problems first problem is

A solar water heater utilizes a pump to lift 400 L of cold water per day through a vertical displacement of 7.0 meters. What is the daily cost of electricity, if the rate is set to be $0.065/kWh? Acceleration due to gravity is 9.81 m/s2 and the density of water is 0.996 g/L.

second problem
Calculate the pressure exerted by the air to the base of the column (area is 1.0 m2). The density of air can be taken a 1.3 kg/m3 and the volume of the container is 100L. Assume ideal gas behaviour and standard conditions (atmospheric pressure and 25 degrees C)

Ok I don't really know where to start with these problems and I'm not looking for the solution to these problems just maybe someone to help me on the right track or maybe to provide me with the equations to do these problems. Thanks in advance.

 

[physics girl phd]

Hint to start you on first problem: how much work do you do on 1 liter of water lifting it that height?

 

[thomate1]

I would be happy to help you with these questions. Let's start with the first problem about the solar water heater. To solve this problem, we need to use the equation P = mgh, where P is the power, m is the mass, g is the acceleration due to gravity, and h is the height or vertical displacement. In this case, we are given the values for m (400 L), g (9.81 m/s2), and h (7.0 meters). We also know that the rate of electricity is $0.065/kWh. To find the daily cost of electricity, we need to first convert the volume of water from liters to kilograms. This can be done by using the density of water, which is 0.996 g/L. Therefore, 400 L of water would be equivalent to 400 kg. Plugging these values into the equation, we get P = (400 kg)(9.81 m/s2)(7.0 meters) = 27,384 J. To convert this to kWh, we divide by 3,600,000 (the number of joules in 1 kWh) and then multiply by the rate of electricity, which gives us a daily cost of $0.065(27,384 J/3,600,000 J) = $0.000495. Therefore, the daily cost of electricity for the solar water heater is $0.000495.

For the second problem, we need to use the ideal gas law, which is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. In this case, we are given the values for V (100 L), the density of air (1.3 kg/m3), and the temperature (25 degrees C or 298 K). We can assume that the number of moles (n) is 1, as we are dealing with a volume of 100 L. We also know that the area of the base is 1.0 m2, so we can use this to find the height (h) of the column of air. Using the formula for volume of a cylinder (V = πr2h), we can rearrange it to solve for h, which gives us h = V/πr2 = (100 L)/(π(0.5 m)2) =