A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 2 years ago
A substance of unknown mass absorbs 138 kilojoules of energy, going from 298 to 303 Kelvin. If the specific heat of the substance is 7.11 J/g·°C, how much of the substance was present?
Answer
3.88 kg
32.3 g
3.88 g
32.3 kg
I need help setting up this equation.
q = mcΔT
138 = 5*(7.11)triangleT
i have this but I feel like its wrong
anonymous
 2 years ago
A substance of unknown mass absorbs 138 kilojoules of energy, going from 298 to 303 Kelvin. If the specific heat of the substance is 7.11 J/g·°C, how much of the substance was present? Answer 3.88 kg 32.3 g 3.88 g 32.3 kg I need help setting up this equation. q = mcΔT 138 = 5*(7.11)triangleT i have this but I feel like its wrong

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0First you need to convert kilojoules to joules 138*1000=138000 J And you must convert kelvins to celsius 298273=25 C 303273=30 Now 3025= 5 So your equation is now 138000=x * 7.11 * 5 138000=35.6x So x=3876g or x=3.88kg D

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks so much for your help!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.