A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 one year ago
An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10°C. Two metallic blocks are placed into the water. One is a 50.0 g piece of copper at 80.0°C; the other block has a mass of 70 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. a) Determine the specific heat of the unknown sample. b) Determine a possible substance for the unknown material
 one year ago
An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10°C. Two metallic blocks are placed into the water. One is a 50.0 g piece of copper at 80.0°C; the other block has a mass of 70 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. a) Determine the specific heat of the unknown sample. b) Determine a possible substance for the unknown material

This Question is Closed

ucreategames
 one year ago
Best ResponseYou've already chosen the best response.0The equation is Q cold = Q Hot 1. Find the specific heat of each Copper (Al) = 900 J/Kg*C Water (H2O) = 4186 J/Kg*C Copper (Cu) = 387 J/Kg*C unknown = x 2. Determine the energy gained or lost. In this problem the water and aluminum gain heat as they were at an equilibrium of 10 C but were raised to 20 C. The Copper was initially at 71 C and our unknown was at 100 C initially but they to ended up at 20 C, so they both lost energy. To do this we find each element separately by using the following formula : Q= MC (TfTi) Q = energy lost or gained, M= mass of the element in kg, C is the specific heat of that element, Tf is the final temp and Ti is the initial temp of that element. Remember to convert from grams to Kg by moving decimal over 3 places to the left. Q (Al) = (.100 Kg) (900 J/Kg*C ) (20*C  10*C) = 900 J/Kg*C (gained) Q (H2O) = (.250 Kg) (4186 J/Kg*C ) (20*C  10*C) =10475 J/Kg*C (gained) Q (Cu) = (.05 Kg) (387 J/Kg*C ) (71*C  20*C) = 987 J/Kg*C (Lost) Q (x) = (.084 Kg) ( x J/Kg*C ) (100*C  20*C) = 6.72x J/Kg*C (Lost) 3. Now apply the formula (Q cold = Q Hot) by adding the numbers together. 900 J/Kg*C + 10475 J/Kg*C = 987 J/Kg*C + 6.72x J/Kg*C We have: 11375 J/Kg*C = 987 J/Kg*C + 6.72x J/Kg*C Subtract 987 from both sides: 11375 J/Kg*C = 987 J/Kg*C + 6.72x J/Kg*C 987 987 10388 J/Kg*C = 6.72x J/Kg*C Divide by sides by 6.72 10388 J/Kg*C = 6.72x J/Kg*C /6.72 /6.72 X= 1545 J/Kg*C Thus the specific heat of our unknown substance is 1545 J/Kg*C

ucreategames
 one year ago
Best ResponseYou've already chosen the best response.0the first is unknown and i forgot the other
Ask your own question
Ask a QuestionFind 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.