The document discusses temperature, heat, and phase changes. It defines temperature as a measure of the average kinetic energy of particles in an object. Temperature scales like Fahrenheit, Celsius, and Kelvin are introduced. Heat is a form of energy that transfers between objects due to temperature differences. Equations are provided to calculate heat transfer during temperature changes and phase changes using values like specific heat and latent heat. Examples demonstrate using the equations to solve heat and temperature problems involving substances like water, ice, and metals.
”Waste heat recovery” is the process of “heat integration”, that is, reusing heat energy that would otherwise be disposed of or simply released into the atmosphere. By recovering waste heat, plants can reduce energy costs and CO2 emissions, while simultaneously increasing energy efficiency.
”Waste heat recovery” is the process of “heat integration”, that is, reusing heat energy that would otherwise be disposed of or simply released into the atmosphere. By recovering waste heat, plants can reduce energy costs and CO2 emissions, while simultaneously increasing energy efficiency.
Ice has been used for hundreds of years for short time preservation of food and still it is used to preserve cool drinks and food for short period. There are many types of ice manufacturing plants. They are divided by thickness and shape of ice. The ice plant consists of four main components condenser, compressor, evaporator and freezing tank containing sodium chloride brine, bring agitator and can with water. In this paper 2000lb per day block ice manufacturing plant is designed. The temperature of water is 27oC and to reduce it to 7.2oC. Based on this requirement, the design consideration and calculation have done. Khaing Zar Nyunt | Hnin Yu Yu Kyaw | Ei Cho Cho Theik ""Design of Ice Manufacturing Plant (2000 lb)"" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-3 | Issue-4 , June 2019, URL: https://www.ijtsrd.com/papers/ijtsrd25115.pdf
Paper URL: https://www.ijtsrd.com/engineering/mechanical-engineering/25115/design-of-ice-manufacturing-plant-2000-lb/khaing-zar-nyunt
Introduction to transient Heat conduction, Lamped System Analysis, Approxiamate Analytical and graphical method and Numerical method for one and two dimensional heat conduction by using Explicit and Implicit method
Ice has been used for hundreds of years for short time preservation of food and still it is used to preserve cool drinks and food for short period. There are many types of ice manufacturing plants. They are divided by thickness and shape of ice. The ice plant consists of four main components condenser, compressor, evaporator and freezing tank containing sodium chloride brine, bring agitator and can with water. In this paper 2000lb per day block ice manufacturing plant is designed. The temperature of water is 27oC and to reduce it to 7.2oC. Based on this requirement, the design consideration and calculation have done. Khaing Zar Nyunt | Hnin Yu Yu Kyaw | Ei Cho Cho Theik ""Design of Ice Manufacturing Plant (2000 lb)"" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-3 | Issue-4 , June 2019, URL: https://www.ijtsrd.com/papers/ijtsrd25115.pdf
Paper URL: https://www.ijtsrd.com/engineering/mechanical-engineering/25115/design-of-ice-manufacturing-plant-2000-lb/khaing-zar-nyunt
Introduction to transient Heat conduction, Lamped System Analysis, Approxiamate Analytical and graphical method and Numerical method for one and two dimensional heat conduction by using Explicit and Implicit method
Discusses macroscopic and microscopic properties of solids and liquids.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
This lesson talks about another gas law, the Charles' Law. This is a simplify lesson for intended for the level of my students understanding. Hope it helps you too!
Younes Sina ,Student Poster Competition ,The Oak Ridge Chapter of ASM, the Ma...Younes Sina
Poster for ASM Oak Ridge Chapter competition Younes Sina, Dr. Carl.J. McHargue
Student Poster Competition (The Oak Ridge Chapter of ASM, the Materials Information Society and the Smoky Mountain Chapter of the Society of Plastic Engineers)
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
A brief information about the SCOP protein database used in bioinformatics.
The Structural Classification of Proteins (SCOP) database is a comprehensive and authoritative resource for the structural and evolutionary relationships of proteins. It provides a detailed and curated classification of protein structures, grouping them into families, superfamilies, and folds based on their structural and sequence similarities.
This pdf is about the Schizophrenia.
For more details visit on YouTube; @SELF-EXPLANATORY;
https://www.youtube.com/channel/UCAiarMZDNhe1A3Rnpr_WkzA/videos
Thanks...!
THE IMPORTANCE OF MARTIAN ATMOSPHERE SAMPLE RETURN.Sérgio Sacani
The return of a sample of near-surface atmosphere from Mars would facilitate answers to several first-order science questions surrounding the formation and evolution of the planet. One of the important aspects of terrestrial planet formation in general is the role that primary atmospheres played in influencing the chemistry and structure of the planets and their antecedents. Studies of the martian atmosphere can be used to investigate the role of a primary atmosphere in its history. Atmosphere samples would also inform our understanding of the near-surface chemistry of the planet, and ultimately the prospects for life. High-precision isotopic analyses of constituent gases are needed to address these questions, requiring that the analyses are made on returned samples rather than in situ.
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
Chloroplasts: Organelles responsible for photosynthesis in plants.
Chloroplast DNA (cpDNA): Circular DNA molecule found in chloroplasts.
Inheritance Pattern: Often maternally inherited in most plants, but can vary in some species.
Examples: Variegation in plants, where leaf color patterns are determined by chloroplast DNA.
Slide 5: Plasmid Inheritance
Plasmids: Small, circular DNA molecules found in bacteria and some eukaryotes.
Features: Can carry antibiotic resistance genes and can be transferred between cells through processes like conjugation.
Significance: Important in biotechnology for gene cloning and genetic engineering.
Slide 6: Mechanisms of Extrachromosomal Inheritance
Non-Mendelian Patterns: Do not follow Mendel’s laws of inheritance.
Cytoplasmic Segregation: During cell division, organelles like mitochondria and chloroplasts are randomly distributed to daughter cells.
Heteroplasmy: Presence of more than one type of organellar genome within a cell, leading to variation in expression.
Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
Petite Mutants in Yeast: Result from mutations in mitochondrial DNA affecting respiration.
Slide 8: Importance of Extrachromosomal Inheritance
Evolution: Provides insight into the evolution of eukaryotic cells.
Medicine: Understanding mitochondrial inheritance helps in diagnosing and treating mitochondrial diseases.
Agriculture: Chloroplast inheritance can be used in plant breeding and genetic modification.
Slide 9: Recent Research and Advances
Gene Editing: Techniques like CRISPR-Cas9 are being used to edit mitochondrial and chloroplast DNA.
Therapies: Development of mitochondrial replacement therapy (MRT) for preventing mitochondrial diseases.
Slide 10: Conclusion
Summary: Extrachromosomal inheritance involves the transmission of genetic material outside the nucleus and plays a crucial role in genetics, medicine, and biotechnology.
Future Directions: Continued research and technological advancements hold promise for new treatments and applications.
Slide 11: Questions and Discussion
Invite Audience: Open the floor for any questions or further discussion on the topic.
Observation of Io’s Resurfacing via Plume Deposition Using Ground-based Adapt...Sérgio Sacani
Since volcanic activity was first discovered on Io from Voyager images in 1979, changes
on Io’s surface have been monitored from both spacecraft and ground-based telescopes.
Here, we present the highest spatial resolution images of Io ever obtained from a groundbased telescope. These images, acquired by the SHARK-VIS instrument on the Large
Binocular Telescope, show evidence of a major resurfacing event on Io’s trailing hemisphere. When compared to the most recent spacecraft images, the SHARK-VIS images
show that a plume deposit from a powerful eruption at Pillan Patera has covered part
of the long-lived Pele plume deposit. Although this type of resurfacing event may be common on Io, few have been detected due to the rarity of spacecraft visits and the previously low spatial resolution available from Earth-based telescopes. The SHARK-VIS instrument ushers in a new era of high resolution imaging of Io’s surface using adaptive
optics at visible wavelengths.
The increased availability of biomedical data, particularly in the public domain, offers the opportunity to better understand human health and to develop effective therapeutics for a wide range of unmet medical needs. However, data scientists remain stymied by the fact that data remain hard to find and to productively reuse because data and their metadata i) are wholly inaccessible, ii) are in non-standard or incompatible representations, iii) do not conform to community standards, and iv) have unclear or highly restricted terms and conditions that preclude legitimate reuse. These limitations require a rethink on data can be made machine and AI-ready - the key motivation behind the FAIR Guiding Principles. Concurrently, while recent efforts have explored the use of deep learning to fuse disparate data into predictive models for a wide range of biomedical applications, these models often fail even when the correct answer is already known, and fail to explain individual predictions in terms that data scientists can appreciate. These limitations suggest that new methods to produce practical artificial intelligence are still needed.
In this talk, I will discuss our work in (1) building an integrative knowledge infrastructure to prepare FAIR and "AI-ready" data and services along with (2) neurosymbolic AI methods to improve the quality of predictions and to generate plausible explanations. Attention is given to standards, platforms, and methods to wrangle knowledge into simple, but effective semantic and latent representations, and to make these available into standards-compliant and discoverable interfaces that can be used in model building, validation, and explanation. Our work, and those of others in the field, creates a baseline for building trustworthy and easy to deploy AI models in biomedicine.
Bio
Dr. Michel Dumontier is the Distinguished Professor of Data Science at Maastricht University, founder and executive director of the Institute of Data Science, and co-founder of the FAIR (Findable, Accessible, Interoperable and Reusable) data principles. His research explores socio-technological approaches for responsible discovery science, which includes collaborative multi-modal knowledge graphs, privacy-preserving distributed data mining, and AI methods for drug discovery and personalized medicine. His work is supported through the Dutch National Research Agenda, the Netherlands Organisation for Scientific Research, Horizon Europe, the European Open Science Cloud, the US National Institutes of Health, and a Marie-Curie Innovative Training Network. He is the editor-in-chief for the journal Data Science and is internationally recognized for his contributions in bioinformatics, biomedical informatics, and semantic technologies including ontologies and linked data.
Seminar of U.V. Spectroscopy by SAMIR PANDASAMIR PANDA
Spectroscopy is a branch of science dealing the study of interaction of electromagnetic radiation with matter.
Ultraviolet-visible spectroscopy refers to absorption spectroscopy or reflect spectroscopy in the UV-VIS spectral region.
Ultraviolet-visible spectroscopy is an analytical method that can measure the amount of light received by the analyte.
2. Temperature
The temperature of an object is a measure of how cold or hot that
object is. More precisely, the temperature of an object is a measure of
the average kinetic energy of the atoms and molecules of that object.
A hotter object has faster molecular vibrations. Temperature is a scalar
quantity.
Temperature Scales
Fahrenheit and Rankin (in the American system of units)
Celsius and Kelvin (in the Metric System)
Fahrenheit and Celsius are regular scales.
Rankin and Kelvin are absolute scales.
3.
4.
5.
6.
7. Example : The Fahrenheit scale reads 77o F, what is the reading on
the Celsius scale?
Solution:
°C = (5/9)[ F - 32 ]
oC = (5/9) [ 77 - 32] = 25
Example : A temperature difference of ΔC = 24oC is measured
between two points on Celsius scale. How much is this difference in
Fahrenheit scale?
Solution: 100oC difference on the Celsius scale corresponds to 180oF
difference on the Fahrenheit scale. Using a proportion, the difference
in Fahrenheit is
ΔF / ΔC = 180o / 100o
ΔF / 24o = 9/5
ΔF = 24o (1.8)
ΔF = 43oF.
C1 = (5/9)[ F1 - 32 ]
C1 = (5/9)[ F1 - 32
C 2 =(5/9)[ F2 - 32 ]
(C1-C2)= (5/9)[(F1-32)-(F2-32)]
ΔC=(5/9)ΔF
8. Example : At what temperature both Fahrenheit and Celsius scales read
the same temperature?
Solution:
°C = (5/9)[ F - 32 ]
F = (5/9)[ F - 32 ]
F=(5/9)F-17.777
(4/9)F=-17.777
F=-40
9. Absolute Scales:
The basis for absolute scales (Kelvin and Rankin scales) is the
temperature at which molecular motion ceases and stops. This
temperature cannot actually be reached; however, with great cooling,
temperatures very close to it have been reached. Experiments have
shown that when a gas is cooled down, its volume decreases.
At constant pressure, the volume decrease for a gas, is proportional to
the temperature decrease. In other words, The ratio ΔV/ΔT remains
constant. That means that if V( the gas volume) is plotted versus T
(the gas temperature) while pressure is kept constant, the graph is a
straight line as shown below:
10. Heat
Heat is a form of energy that transfers due to a temperature difference.
Units of Heat
The familiar unit often heard is "calorie."
One calorie (1 cal) is the amount of heat energy that can raise the temperature
of 1 gram of pure water by 1 oC.
Parallel to this definition is that of kilocalorie (kcal).
1 kcal is the amount of heat energy that can raise the temperature of 1 kg of
pure water by 1 oC.
A non-Metric unit for heat energy is Btu (British thermal unit).
1 Btu is the amount of heat energy that can raise the temperature of 1 lbm of
pure water by 1 oF.
11. Specific Heat (c)
Different substances take different amounts of heat energy for one unit
of mass of them to warm up by one degree. For example, if you place
1 kg of copper and 1 kg of aluminum on a burner, after the same length
of time, the copper piece will be much hotter than the aluminum. The
reason is the difference in their specific heat. Copper takes much less
heat to warm up compared to aluminum.
12. The specific heat (c) of a substance is the amount of heat that 1kg of
substance takes to warm up by 1oC. On this basis, the specific heat of
water is 1 kcal /(kg oC). The specific heat of a few elements are given
below:
Cwater =1.000 kcal / ( kg oC) or, cwater = 1.000 cal /(g oC)
ciron = 0.108 kcal / ( kg oC) or, ciron = 0.108 cal /(g oC)
cAl = 0.215 kcal / ( kg oC) or, cAl = 0.215 cal /(g oC)
Cbrass = 0.0924 kcal / ( kg oC) or, cbrass= 0.0924 cal /(g oC)
13. Heat calculation:
When heat is given to a pure substance or taken from it, its
temperature starts changing if a phase change does not start. During
a phase change (solid to liquid, liquid to solid, liquid to vapor, or
vapor to liquid ),temperature remains constant. We will look at the
following two cases:
1) Heat calculation for temperature change
2) Heat calculation for phase change
14. Heat Calculation When Temperature changes
Q = McΔT
No Phase Change
Mass of the object
specific heat
temperature change
15. Example :
Calculate the amount heat that must be given to 2.14 kg of iron to
warm up from 24.0 oC to 88.0 oC.
Solution:
Q = McΔT
Q = ( 2.14kg )[0.108 kcal /(kg oC)] ( 88 - 24 )oC = 14.8 kcal
ΔE
16. Example :
37.0 cal of heat is given to 2.00 gram of water at 12.0 oC. Find its
final temperature.
Solution:
Q = McΔT
ΔT = Q / Mc
ΔT = (37cal) / [(2gr)(1cal/ gr oC)] = 18.5oC
Tf - Ti = 18.5 °C
Tf - 12.0 oC = 18.5 oC
Tf = 30.5 oC
Example :
Calculate the amount heat that must be taken from 5.00 kg of
Aluminum to cool it down from 230oC to 30oC.
Solution:
Q = McΔT
Q =(5.00kg )[0.215 kcal / ( kg oC )](30 - 230 )oC = - 215 kcal
18. Example :
A 65-gram piece of aluminum at 180 oC is removed from a stove and
placed in 45 grams of water initially at 22 oC. Find the equilibrium
temperature (Teq).
19. Solution:
-[ heat loss by hotter objects ] = heat gain by colder objects
Q = McΔT
-Mc [Tf - Ti] Al = Mc[Tf - Ti] water
- (65)(0.215)(Teq - 180) = (45)(1.00)(Teq - 22)
-14(Teq - 180) = (45)(Teq - 22)
-14Teq + 2520 = 45Teq - 990
3510 = 59Teq
Teq = 59 oC
20. Example :
A 225-gram piece of hot iron is removed from an electric oven at an
unknown temperature. It is known that the iron piece has been in the
oven long enough so that its initial temperature can be thought as the
temperature of the oven. The iron piece is placed in 75.0 grams of water
that is held by a 45.0-gram aluminum container initially at an equilibrium
temperature of 25.0 oC. The final equilibrium temperature of iron,
aluminum, and water becomes 41.0 oC. Find the initial temperature
of iron (oven). Assume that the whole system is thermally isolated
from the surroundings.
hot iron
225 gram
75.0 gram
water
45.0 gram
aluminum
25.0 oC
41.0 oC
21. Solution:
-[ heat loss by hotter objects ] = heat gain by colder objects
- Mc [Tf - Ti]iron = Mc [Tf - Ti]Al + Mc [Tf - Ti]water
-(225)(0.108)(41-Ti ) = (45)(0.215)(41-25) + (75)(1)(41-25)
-24.3(41-Ti ) = 1354.8
-41-Ti = 55.8
Ti = 96.8 oC
22. Heat Calculation
Phase Change Only
During a phase change such as following, temperature remains constant:
solid to liquid
liquid to solid
liquid to vapor
vapor to liquid
23. Q = M Lf
latent heat of fusion
Lf is measured for different substances at their melting/freezing points or
temperatures. Typical values may be found in texts or handbooks.
24. Example : How much heat should be removed from 250 grams of water
already at its freezing point (0 oC) to convert it to ice at (0 oC)? The latent
heat of fusion (freezing) for water is Lf = 80 cal/gr.
Solution: In this problem, the heat calculation involves a phase change
only. There is no temperature change.
Q = M Lf
Q = (250gr)(80 cal/gr ) = 20,000 cal
0 oC 0 oC
25. Example :
How much heat should be given to 250 gr of ice at 0oC to convert it to water
at 40.oC ? The latent heat of freezing/melting for water is Lf = 80 cal /gr and
the specific heat of water is 1 cal/[gr oC].
Solution: In this problem, the heat calculation involves a phase change and
a temperature change.
Q = M Lf + Mc (Tf - Ti)
Q = (250)(80) + (250)(1)(40 - 0) = 30,000cal
26. -20 oC
0 oC
250 gr
100 oC
Example :
How much heat should be given to 250 gr of ice at -20 oC to bring it to boil (100 oC)?
0 oC
27. Solution:
This problem has 3 steps:
A temperature change ( ice from -20oC to 0oC )
A phase change ( ice at 0oC to water at 0°C )
A temperature change ( water from 0oC to water at 100oC)
cice = 0.48 cal / (gr oC)
Lf = 80 cal/gr
cwater = 1cal / (gr oC)
Q = Mcice (Tf - Ti) + M Lf + Mcwater (Tf - Ti)
Q = (250)(0.48)[ 0 - (-20) ] + (250)( 80) + (250)( 100 - 0) = 47,000 cal
2400cal 20,000cal 25,000cal
phase changetemperature change temperature change
28. Example :
Draw a diagram that shows temperature change vs. heat consumed for previous Example.
Solution:
29. Example :
2650 cal of heat is given to 225 grams of ice at -12.0 oC. Find the final
temperature of the result.
Solution: All of the ice may not melt. Let's first calculate the heat necessary
for the ice to warm up to 0 oC-ice.
Q = Mc ( Tf - Ti ) = (225gr)(0.480 cal/gr oC)[ 0 - (-12.0)]oC = 1300 cal
The remaining heat is 2650cal - 1300cal = 1350 cal
Now, let's see if this heat is enough to melt the ice
Q = MLf = (225gr)(80. cal /gr ) = 18000 cal
No, 1350 cal is not enough to melt all of the ice. We need to see how much of
the ice does melt.
Q = MLf
1350cal = M (80 cal / gr)
M = 17 grams.
The final result is a mixture of
17 grams of water and (225 - 17) grams
of ice, of course both at 0oC.
30. Example :
26500 cal of heat is given to 225 grams of ice at -12.0 oC. Find the final temperature.
Solution:
All of the ice may not melt. First calculate the needed heat for the ice to warm up
to 0 oC-ice.
Q = Mc ( Tf - Ti ) = (225gr)(0.480cal /(gr oC)[ 0.0 - (-12.0)]oC = 1300 cal
The remaining heat is 26500cal - 1300cal = 25200 cal
Now, let's see if this heat is enough to melt the ice.
Q = MLf = (225gr)(80cal /gr ) = 18000 cal
Yes, 25200 cal is enough to melt all of the ice.
The remaining heat is 25200cal - 18000cal = 7200 cal. This heat warms up water
from 0oC to Tf .
Q = Mc ( Tf - Ti )
7200cal = (225gr)(1 cal/gr oC)( Tf - 0oC )
Tf - 0.0oC = 7200 / 225 = 32oC
Tf = 32oC
The final result is 225 grams of water at 32oC.