Question

I force 200 cm³ of water through a slanted pipe whose cross-sectional area is constant at 10 cm² . the pressure at the pipe inlet is 150 n/cm² higher than at its outlet. how much work have i done on the system?

a. 150 J
b. 300 J
c. 3000J
d. It depends on the height difference at the ends of the pipe

Answer 1

The work done on the system by forcing 200 cm³ of water through a pipe with a constant cross-sectional area and a pressure difference of 150 N/cm² is 30,000 N·cm or 300 joules.

Step-by-step explanation:

To calculate the work done on the system by forcing 200 cm³ of water through a pipe with a pressure difference of 150 N/cm², we can use the formula W = PΔV, where W is the work, P is the pressure difference, and ΔV is the change in volume. Since the volume of water pushed is 200 cm³ and the pressure difference is 150 N/cm², the work done is W = 150 N/cm² × 200 cm³. The work done is therefore 30,000 N·cm or 300 J (joules), because 1 N·cm = 0.01 J.

It is important to note that this calculation assumes the cross-sectional area of the pipe remains constant and does not take into account any height difference or gravitational potential energy work that might be involved in the fluid's movement. For a real-world application, these factors might need to be considered for a more accurate calculation.