Watching paint dry is boring, so instead you drop a rock into a bucket of water.
You take a 2.5 kg rock and bucket filled with 10 kg of water as your closed thermodynamic system. Everything (rock, bucket, water, environment) starts out at the same temperature. You position the rock 3 m above the water, then drop it into the bucket. (No water is lost from the bucket.)
After the rock has been dropped into the water and the system has reached equilibrium at its original temperature...
How does all the energy (internal ΔU, kinetic ΔKE, potential ΔPE, heat flow Q, work done W) change?
- A) ΔU = 0 J; ΔKE = 0 J; ΔPE = -73.5 J; Q = -73.5 J; W = 0 J
- B) ΔU = 73.5 J; ΔKE = 0 J; ΔPE = -73.5 J; Q = 0 J; W = 0 J
- C) ΔU = 0 J; ΔKE = 73.5 J; ΔPE = -73.5 J; Q = 0 J; W = 0 J
- D) ΔU = -73.5 J; ΔKE = 0 J; ΔPE = -73.5 J; Q = -73.5 J; W = 0 J
- Not sure, Sherlock