Fox and McDonald's Introduction to Fluid Mechanics
9th Edition
ISBN: 9781118912652
Author: Philip J. Pritchard, John W. Mitchell
Publisher: WILEY

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Textbook Question
Chapter 4, Problem 1P

An ice-cube tray containing 250 mL of freshwater at 15°C is placed in a freezer at −5°C. Determine the change in internal energy (kJ) and entropy (kJ/K) of the water when it has frozen.

To determine

The change in internal energy and entropy of the water.

### Explanation of Solution

Given:

The volume of the water (V) is 250 ml.

The temperature of the water (T1) is 15°C.

The temperature of the freezer (T2) is 5°C.

Calculation:

Consider the density of the water (ρ) is 999kg/m3 and the specific heat at constant volume (cP) is 1 kcalkgK.

Calculate the change in entropy (ΔS).

ΔS=m(s2s1)ΔS=ρVcvln(T2T1)=999kg/m3×(250ml)×1 kcalkgK×ln(5°C15°C)=999kg/m3×(250ml)×1 kcalkgK×ln(268 K288 K)=0.01797kcalkg×4190 J1 kcal=75.32 J/K=0.0753 kJ/K

Thus, the change in entropy is (Δs) is 0.0753 kJ/K_.

Calculate the change in internal energy (ΔU).

ΔU=mcv(T2T1)ΔU=ρVcv(T2T1)=999kg/m3×(250ml×106m31ml)×1 kcalkgK×(5°C15°C)=999kg/m3×(250×106m3)×kcalkgK×(268 K288 K)=4.995kcalkg×4190 J1 kcal=20929.05 J/K=20.929 kJ/K

Thus, the change in internal energy is (Δs) is 20.929 kJ/K_.

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