The units supplied for 130 indicate specific heat capacity, so the block's total heat capacity is

C = m c = 0.5 (130) = 65 J K^{- 1}

. Treat it as constant, and the lake as a thermal reservoir at

T_{0} = 283.15 K

. The initial block temperature is

T_{i} = 373.15 K

Compute the block's entropy change using a reversible comparison path, even though the actual cooling is irreversible:

Δ S_{block} = \int_{T_{i}}^{T_{0}} \frac{C d T}{T} = C ln \frac{T _{0}}{T _{i}} = - 17.940 J K^{- 1} .

(1)

The lake receives

Q = C (T_{i} - T_{0}) = 5850 J

at constant temperature, so

Δ S_{lake} = \frac{Q}{T _{0}} = 20.660 J K^{- 1} .

(2)

Therefore

Δ S_{Universe} = 2.72 J K^{- 1} .

(3)

This is entropy generated by cooling a finite body in a reservoir. The positive total is consistent with the Second law of thermodynamics, although the cooling block's own entropy decreases.

Initially and finally the block and lake are at the same temperature. After the falling block has come to rest, its lost gravitational potential energy has dissipated into internal energy of the lake and surroundings. In the idealization that all

m g h

is ultimately thermalized in the lake at

T_{0} = 283.15 K

Q = m g h = (0.5) (9.81) (200) = 981 J .

(1)

The block returns to its original thermal state, so its net entropy change is zero. Mechanical potential energy carries no thermal entropy that must be subtracted. Thus

Δ S_{Universe} = \frac{m g h}{T _{0}} = 3.46 J K^{- 1} .

(2)

Use of

g = 9.80 m s^{- 2}

gives the same three-significant-figure answer. Splashing and any temporary heating do not change this final-state balance if their mechanical energy is eventually dissipated in the same reservoir.