Hi folks, please help me with this hw problem. Thanks. (Modified Olander Problem

Question

Hi folks, please help me with this hw problem. Thanks.

(Modified Olander Problem 2.3) A 10 kg specimen of copper is quenched at constant pressure from 700 °C in an adiabatic, 100 liter bath of water that is initially at 20 °C. Neglecting the vaporization of the water (assume it all stays liquid) and the heat capacity of the bath walls, calculate the final temperature of the system using Figure 2.7 for the heat capacity of copper and the Lav of Dulong and Petit for the heat capacity of the water. Use Cp since the quenching occurs at constant pressure, and remember to multiply the Law by the number of atoms per molecule! This problem can be useful for vou materials scientists who want to design a quenching process for laboratory samples: If the final temperature will be near 100 °C, add more water to the bath to keep it from boiling over.

Explanation / Answer

Cp of copper (700 c) = 26 J/mol-K

molecular wieght of copper = 63

heat capacity of 10 kg Copper = 10*700*26/0.063 = 2.89e+6 J

Molar heat capcity of water = 3R = 25 J/mol/K

molecular weight of water = 18 g

inital temp =20.

Water 100 Lts = 100 kg

Total heat capacity of water = (100/0.018) *20 *25 = 2.78e+6 J

Let T be the final temp of the bath then

heat last by copper = 2.89e+6 - 10*TCp /0.063 = 2.89e+6 - 158.73 T Cp

heat gained by water = (T-20)(100/0.018) *25   = 1.39e+5 T - 2.78e+6 J

heat lost by copper = heat gained by water

2.89e+6 - 158.73 T Cp   = 1.39e+5 T - 2.78e+6 J

(1.39e+5 + 158.73 Cp) T = 5.67 e+6

at low temp ,100 Cp is very small and the contribution of the second term on LHS is negligible and can be ignored for simplification

T = 5.67e+6/1.39e+5 = 40.79 C, final temp

at this temp heat capcity of copper = 10*40*5/0.063 = 31746 J , small compred to the heat capacity of water

and hecne our assumption holds good

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