Answer:
0.167 kJ
Explanation:
Given:
Volume of oil delivered by the pump (V) = 0.628 dm³ = 628 cm³ [1 dm³ = 1000 cm³]
Density of the oil (d) = 0.776 g/cm³
Acceleration due to gravity (g) = 9.8 m/s²
Time for which pump runs (t) = 41 s
Height to which oil is raised (h) = 35 m
Now, we know that, mass of a body is equal to the product of its volume and density. Therefore,
mass of oil (m) = volume × density
[tex]m=628\ cm^3\times 0.776\ g/cm^3\\\\m=487.328\ g=0.487\ kg\ \ \ \ \ \ \ [1\ g=0.001\ kg][/tex]
Now, work done by the pump is equal to the increase in the potential energy of the oil being raised to the given height.
Increase in potential energy is given as:
[tex]\Delta U=mgh[/tex]
Therefore, the work done by the pump is given as:
Work done = ΔU
[tex]W=mgh\\\\W=(0.487\ kg)\times (9.8\ m/s^2)\times (35\ m)\\\\W=167.04\ J[/tex]
Now, 1 J = 0.001 kJ
Therefore, 167.04 J = 167.04 × 0.001 = 0.167 kJ
Hence, work done by the pump in kilojoules is 0.167 kJ
