A 0.750-kg iron bar is taken from a forge at 900 $\circ$ C and dropped into 14.00 kg of water at 20.0 $\circ$ C.Assuming that no energy is lost by heat to the surroundings as the water/iron reaches their final temperature,determine the total entropy change of the iron bar plus-water system.Assume that specific heat of iron is 448 J/(kg.K).

Question

Quizplus · Accepted Answer

The entropy change of the system can be calculated using:

ΔS = ΔSiron + ΔSwater

where ΔSiron = mCln(final temperature/initial temperature) and ΔSwater = mCln(final temperature/initial temperature), with m being the mass and C being the specific heat capacity of each substance.

First, let's calculate the final temperature of the system. Using the heat lost by the iron bar being equal to the heat gained by the water:

mironCironΔT = -mwaterCwaterΔT

(0.750 kg)(448 J/(kg.K))(Tfinal - 900°F) = -(14.00 kg)(4186 J/(kg.K))(Tfinal - 20°F)

Solving for Tfinal, we get:

Tfinal = 25.7°C = 77.3°F

Now we can calculate the entropy changes:

ΔSiron = (0.750 kg)(448 J/(kg.K))ln(25.7°C/232°C) = -187 J/K
ΔSwater = (14.00 kg)(4186 J/(kg.K))ln(25.7°C/20°C) = 721 J/K

ΔS = -187 J/K + 721 J/K = 534 J/K

Therefore, the best choice is C.

A 0750-Kg Iron Bar Is Taken from a Forge at 900