9. Will took a 45.0g cube of solid H2O (an ice cube) out of the freezer at -5.00°C and put it on the table inthe lab. The ice cube melts by absorbing heat from the air in the room, which changes slightly, but to foursignificant figures is at a constant 20.00°C, and the melt-water reaches room temperature.a) Start with AS = S to find the total change in entropy for the H2O for this entire process.b) Start with AS = S to find the change in entropy for the air for this entire process.c) Find the total change in entropy for the universe for this process.d) Use the answer from part c) to explain whether or not this process is reversible.

Answered: 9. Will took a 45.0g cube of solid H20…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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9. Will took a 45.0g cube of solid H2O (an ice cube) out of the freezer at -5.00°C and put it on the table in
the lab. The ice cube melts by absorbing heat from the air in the room, which changes slightly, but to four
significant figures is at a constant 20.00°C, and the melt-water reaches room temperature.
a) Start with AS = S to find the total change in entropy for the H2O for this entire process.
b) Start with AS = S to find the change in entropy for the air for this entire process.
c) Find the total change in entropy for the universe for this process.
d) Use the answer from part c) to explain whether or not this process is reversible.

Expert Solution

Step 1

Given data,

m=45 g

$θ$ ₁ = -5°C

$θ$ = 0°C

$θ_{2}$ = 20⁰C

L_ice = 8o cal / g,

S_ice = 1/2 cal/g ⁰C

S_water = 1 cal/g

Q_total= Q₁ +Q₂ +Q₃

(a) $∆ S = \int \frac{d Q}{T} = \int (m S_{i c e} \frac{d T}{T} + \frac{m L}{T} + m S_{w a t e r} \frac{d T}{T}) = m S_{i c e} \int_{θ_{i}}^{θ} \frac{d T}{T} + \frac{m L}{T} + m S_{w a t e r} \int_{θ}^{θ_{2}} \frac{d T}{T} = 45 \times \frac{1}{2} {[\ln T]}_{268 K}^{273 K} + \frac{45 \times 80}{273} + 45 \times 1 {[\ln T]}_{273}^{293} = 22.5 \ln (\frac{273}{268}) + \frac{45 \times 80}{273} + 45 \times \ln (\frac{293}{273}) = 16.78 \frac{c a l}{K m o l}$

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