A, Inside a calorimeter is 100 g of water at 39.8 ºC. A 10 g object at 50 ºC is placed inside of a calorimeter. When equilibrium has been reached the new temperature of the water and metal object is 40 ºC. What type of metal is the object made from? Mass of water = 100 g Specific heat of water = 4.186 j/g Change in temp of water = .2 ºC Mass of metal = 10 g Specific heat of metal = unknown Change in temp of the metal = 10 ºC Mass of water x SH of water (Change in temp of water) = Mass of metal x SH of metal (Change in temp of metal) 100g x 4.186 j/g( .2) = 10g x SH of metal( 10) 83.72 =100 =.8372 j/g Why is ΔTmetal < 0? Why is ΔTwater > 0 ?
A, Inside a calorimeter is 100 g of water at 39.8 ºC. A 10 g object at 50 ºC is placed inside of a calorimeter. When equilibrium has been reached the new temperature of the water and metal object is 40 ºC. What type of metal is the object made from? Mass of water = 100 g Specific heat of water = 4.186 j/g Change in temp of water = .2 ºC Mass of metal = 10 g Specific heat of metal = unknown Change in temp of the metal = 10 ºC Mass of water x SH of water (Change in temp of water) = Mass of metal x SH of metal (Change in temp of metal) 100g x 4.186 j/g( .2) = 10g x SH of metal( 10) 83.72 =100 =.8372 j/g Why is ΔTmetal < 0? Why is ΔTwater > 0 ?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I need help with question number 1 and 2 at the end.
I attempted question A and would like to know if i did it right
thanks so much
A, Inside a calorimeter is 100 g of water at 39.8 ºC. A 10 g object at 50 ºC is placed inside of a calorimeter. When equilibrium has been reached the new temperature of the water and metal object is 40 ºC. What type of metal is the object made from?
Mass of water = 100 g
Specific heat of water = 4.186 j/g
Change in temp of water = .2 ºC
Mass of metal = 10 g
Specific heat of metal = unknown
Change in temp of the metal = 10 ºC
Mass of water x SH of water (Change in temp of water) = Mass of metal x SH of metal (Change in temp of metal)
100g x 4.186 j/g( .2) = 10g x SH of metal( 10)
83.72 =100
=.8372 j/g
- Why is ΔTmetal < 0?
- Why is ΔTwater > 0 ?
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