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Burka [1]
3 years ago
14

Part A. Two containers, one at 305 K and the other at 295 K, are placed in contact with each other. 1. 1 J of heat flows from th

e hot container to the cold container. Find the change in entropy of the hot container and the cold container, and of the whole system (hot cold container together). Does this process violate the second law of thermodynamics
Chemistry
1 answer:
posledela3 years ago
7 0

Answer:

0.00011 JK.

The process does NOT violate the second law of thermodynamics

Explanation:

The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.

=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.

The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.

The change in entropy of the cold container = 1/295 = 0.00339 J/K.

Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.

Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.

The process does NOT violate the second law of thermodynamics.

You might be interested in
What is the volume of 40.0 grams of argon gas at STP ?
MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

  • States of Matter

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

24.9235 L Ar ≈ 24.9 L Ar

5 0
3 years ago
How much energy is used to melt 44.33 g of solid oxygen?
Nutka1998 [239]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

4 0
2 years ago
Read 2 more answers
Urgent!! A chemist measured 5.2 g copper(II) bromide tetrahydrate (CuBr2•4(H2O)). How many moles were measured out? Answer in un
bezimeni [28]
  The  moles  which  were   measured  out  is  calculated  using  the  following  formula

moles  =  mass/molar   mass

molar mass  of  CuBr2.4H20  =   63.5  Cu + (  2  x79.9)  br  + ( 18  x4_)  h20  =  295.3  g/mol

moles  is therefore=  5.2 g/  295.3 g/mol=  0.0176 moles
4 0
3 years ago
Read 2 more answers
How many grams of NaCl are needed to prepare 60 g of a 6.0% solution?
AlekseyPX

Answer:

3.6 grams of NaCl are needed

Explanation:

Percent solution are solutions whose concentrations are expressed in percentages. The amount(either weight or volume) of a solute is expressed as a percentage of the total weight or volume of solution. Percent solutions can either be expressed as  weight/volume % (wt/vol % or w/v %), weight/weight % (wt/wt % or w/w %), or volume/volume % (vol/vol % or v/v %).

A 6.0% wt/wt % solution contains 6 g of solute in 100 g of solution

Therefore, a 100 g solution contains 6.0 g of solute.

60 g of 6.0% solution will contain  60 g solution * 6.0 g solute/ 100 g solution

Mass of NaCl present =  3.6 g  of NaCl

3 0
3 years ago
Calculate the energy for the transition of an electron from the n = 5 level to the n = 6 level of a hydrogen atom. E = Joules Is
kramer

Answer:

For an electron to move from a lower energy level to a higher energy , that electron needs to absorb energy sufficient enough to excite it to make the transition. Hence it is an absorption process. The required energy of transition  E = 2.665 x 10⁻²⁰J

Explanation:

Using the Rydberg's equation we can calculate the wavelength of the photon of energy transition as follows:

1/λ = R . (1/nf² - 1/ni²)

where

λ is the required wavelength of the photon needed to be absorbed to excite the electron to transit from level 5 to 6.  

(Note that for the electron to transit to from energy level 5 to 6, the photon would have to fall from level 6 to 5 in order to emit the required energy to excite the electron)

R is the Rydberg's constant 1.097 x 10⁷ m⁻¹

nf is the final level of the photon

ni is the initial level of the photon

1/λ = 1.097 x 10⁷ m⁻¹ (1/5² - 1/6²)

1/λ = 1.3407 x 10⁵ m⁻¹

λ = 7.458 x 10⁻⁶ m

This implies that that is the wavelength of the photon required to excite the electron to transit from energy level 5 to 6. Using the equation below, we can calculate the energy of transition as

E = h.c/λ

where

E is the required energy of transition

h is the Planck's constant (6.626 x 10⁻³⁴ Js)

c is the speed of light (3 x 10⁸ms⁻¹)

λ is the wavelength calculated above

E = 6.626 x 10⁻³⁴ Js  x  3 x 10⁸ms⁻¹/ 7.458 x 10⁻⁶ m

E = 2.665 x 10⁻²⁰J

3 0
3 years ago
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