# Example Problem with Complete Solution

 4C-1 : Application of the 1st Law to a Stone Falling Into Water 5 pts Consider a stone having a mass of 10 kg and a bucket containing 100 kg of liquid water.  Initially, the stone is 10.2 m above the water and the stone and the water are at the same temperature, state1.  Determine ΔU, ΔEkin, ΔEpot, Q and W for the following changes of state, assuming g = 9.8066 m/s2. a.) From state 1 until the stone is about to enter the water, state 2. b.) From state 2 until the stone has just come to rest at the bottom of the bucket, state 3. c.) From state 3 until heat has been transferred to the surroundings in such an amount that the stone and water in state 4 have returned to their initial temperature, T4 = T1. Read : Choose the combination of the stone and the water as the system. In step 1-2, if no friction or heat transfer exist, potential energy is converted into kinetic energy. In step 2-3, if the water has negligible depth, kinetic energy is converted into internal energy by friction between the stone and the water. In step 3-4, heat transfer from the system to the surroundings reduces the internal energy of the system back to its initial value. Given: T1 100 Find: T2 600 ΔU ??? kJ h1 10.2 m ΔEkin ??? kJ mw 100.0 kg ΔEpot ??? kJ ms 10.0 kg Q ??? kJ Diagram : Assumptions: 1 - Friction between the air and the stone is negligible. 2 - The air and the stone are at the same temperature. 3 - The depth of the water is very small, compared to h1. Equations / Data / Solve : The starting point for this problem is the integral form of the1st Law : Eqn 1 Part a.) As the stone falls through the air, it experiences some air friction, but we can assume that this is negligible.  Consequently, there is no change in the temperature or internal energy of the stone.  If we further assume that the air and stone are at the same temperature, then no heat transfer occurs during this step.  Finally, if we consider the stone and the water to be our system, then no work has crosses the system boundary either. Eqn 2 Eqn 3 Eqn 4 This allows us to simplify the 1st Law to : Eqn 5 Next, we can evaluate ΔEpot from its definition. Eqn 6 When we apply this equation to our problem, Δz = h1 = -10.2 m.  So we can now plug values into Eqn 3. g 9.8066 m/s2 ΔEpot -1000.3 J gc 1 kg-m/N-s2 Now, we can use Eqn 2 to evaluate ΔEkin : ΔEkin 1000.3 J Part b.) Apply the 1st Law, Eqn 1,  to a process from State 2 to State 3, again using the stone and the water as our system. Assume that the depth of the water is negligible so that: Eqn 7 Because the water and stone are at the same temperature, no heat transfer occurs, therefore : Eqn 8 Just as in part (a), no work crosses the boundary of the system (the stone and the water) : Eqn 9 Now, use Eqns 7 - 9 to simplify the 1st Law, Eqn 1 to : Eqn 10 Eqn 10 tells us that in step 2-3, all of the kinetic energy of the stone is converted into internal energy in both the stone and the water. Since the kinetic energy of the stone in state 1 is zero: Ekin,1 0 J We conclude from part (a) that: Ekin,2 1000.3 J After the stone hits the bottom of the tank, it has zero kinetic energy: Ekin,3 0 J Therefore, for step 2-3: ΔEkin -1000.3 J Plug this value of ΔEkin into Eqn 10 to get: ΔU 1000.3 J Part c.) Apply the 1st Law, Eqn 1,  to a process from State 3 to State 4, again using the stone and the water as our system. In step 3-4, there is no change in either the kinetic or the potential energy of the system.  No work crosses the boundary of the system.  Therefore : Eqn 11 Eqn 12 Eqn 13 This allows us to simplify the 1st Law to : Eqn 14 Because in step 3-4 the system returns to its original temperature: Eqn 15 ΔU -1000.3 J Finally, we can plug this value for ΔU back into Eqn 14 to evaluate Q34: Q34 -1000.3 J Note that the negative value for Q34 means that heat is transferred from the system to the surroundings. Part d.) We can determine the values Q, W, ΔU, ΔEkin and ΔEpot for the process from state 1 to state 4 by adding the results from parts (a) through (c). Q14 -1000.27 J W14 0.0 J ΔU 0.0 J ΔEkin 0.0 J ΔEkin -1000.27 J Verify: The assumptions made in this problem solution cannot be verified. Answers : Q (J) W (J) ΔU (J) ΔEkin (J) ΔEpot (J) a.) 1-2 0 0 0 1000 -1000 b.) 2-3 0 0 1000 -1000 0 c.) 3-4 -1000 0 -1000 0 0 d.) 1-4 -1000 0 0 0 -1000.3 