STEM Editing and Review (Contract)

Join our team of proficient and professional reviewers who are dedicated to helping students learn and understand academic concepts. Study.com looking for knowledgeable and experienced experts in STEM fields to help ensure accuracy and clarity in our example problems, practice question sets, and answers to difficult academic homework questions. We recently ranked 10th on Virtual Vocations' top 25 list of best partners for remote work in 2021, and we're growing our team of contractors even further!

What we're looking for:

Expert knowledge, education, and/or experience in a STEM field
Excellent proofreading and editing skills
Strong English reading and writing fluency
Teaching and/or tutoring experience (not required, but preferred)

Our immediate needs exist in the following disciplines, though we have many other STEM opportunities available as well: Math, Chemistry, Physics

What we offer:

Reliable payments twice a month and automated invoicing for your work
Fully online contracted work-from-home opportunity
Flexible work requirements (Basically, no requirements! Work when you want, where you want, as often as you want, with no minimums/maximums)
Supportive staff to answer your questions and help you get up and running

About Study.com

At Study.com, our quest is to make education fun and accessible. We're working to help students by answering their tough homework questions with high quality information to guide them on their academic journey. If you're passionate about helping to make education accessible, apply today!

You can view a step-by-step overview of the contract process here: How to Become a Freelancer with Study.com

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I feel confident reviewing academic questions and sample problems in the following fields* High School MathematicsCollege-Level MathematicsPhysicsStatisticsBiologyChemistryGeneral Science

Please describe your qualifications to review and edit difficult homework questions in your chosen subject(s). Write exactly 3 sentences.*

Reviewing academic STEM answers for Study.com often involve understanding and using LaTeX to format equations and solutions. LaTeX follows simple guidelines, and makes it easy to create beautifully formatted solutions. If you don't know LaTeX, that's OK. But, you will need to acquire at least a basic understanding in order to review Expert Answers. There are many good online resources to learn about LaTeX if you are not currently familiar.

Please edit the following incorrect LaTeX expression:

5\square{3x} + 2x**2 – \frac{3x/2}*

Read the answer below. Check for grammar, accuracy, plagiarism, etc., then provide the writer of this answer with specific feedback that will help them understand what they need to fix.
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Question:
Suppose you have 600.0 grams of room temperature water (20.0 degrees Celsius) in a thermos. You drop 90.0 grams of ice at 0.00 degrees Celsius into the thermos and shut the lid.
(a) What is the equilibrium temperature of the system?
(b) How much ice is left (in grams)?

Answer:

a. the equilibrium temperature of the system is 7.74 celsius
b. There will be 0 grams of ice left

The problem can be solved by breaking it into smaller pieces. The amount of energy needed to melt the ice is Wi×Hf, where Wi is the weight of the ice. In addition, the 90 grams of water obtained from the ice must be heated from 0°C, which is the temperature of the ice, to the final temperature Tf.

Where does the energy for melting the ice come from? Since the cup is insulated, the heat has to come from the 600 grams of water at 20°C. The amount of heat required to melt the ice and to increase the temperature of the resulting water to the final temperature is equal to the heat lost by the 100 grams of water. This can be expressed by an equation:

Wi×Hf + Wi×Sh×(Tf - 0°C) = Ww×Sh×(Twi - Tf)
Where
Wi = weight of ice = 90g
Hf = Heat of fusion = 80 calories/gram
Ww = weight of water = 600g
Tf = final temperature,
Twi = initial water temperature = 20°C
Sh = Specific Heat for water = 1.0 calories per gram per °C
(90×80) + (20×1)×(Tf - 0°C) = (600×1)×(20°C - Tf)
7200 + 20Tf = 12000 - 600Tf
620Tf = 4800
Tf = 4800/620 = 7.74°C
Using the equation above, where the final temperature is 0°C
and the weight of the ice is unknown, we have:
Wi×80 = 600×1×20
The maximum amount of ice that could be melted is:
Wi = 12000/80 = 150 grams
Then, use the initial mass of ice – amount melted
= 150 g – 150g = 0 grams of ice*

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STEM Expert Reviewer (Contract)

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