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ID: 586263 • Letter: I
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i need help with this lab please https://m.youtube.com/watch?feature=youtu.be&v=1iMzFIL25IU
Data Experiment 9; Calorimetry A. Specific Heat Trial 1 Trial 2 Mass of unknown metal sample Mass of calorimeter Mass of calorimeter and water Mass of water Initial temperature of water in calorimeter Initial temperature of metal Equilibrium temperature of metal and water 4.967g 3.869g 52.785g 5.249g 3.869g 52.852g 21.0 C 95.2 C 22.5 C 21.1C 95.2C 22.7C Tmetal-final . Tiatal Specific heat of the metal (Eq. 3) Average Specific heat for the reaction B. Heat of Solution Mass of ionic solid, ms. Mass of calorimeter plus water Mass of water, mico Original temperature of water Final temperature of solution quo for the reaction (Eq1(-418c) 4.129g 4.105g 52.446g 53.046g 21.0C 21.0C 16.5C 16.7CExplanation / Answer
A specific heat
For Trial -1
Mass of water = Mass of calorimeter and water - Mass of calorimeter
= 52.785g – 3.869g = 48.916 g
TH2O =Tfinal – T initial
= 22.5 – 21 = 1.5oC
Tmetal =Tfinal – T initial
= 95.2 – 22.5 = 72.7oC
q Water = mc TH2O
c = specific heat capacity of water = 4.184 J/g 0C
q Water = m c t
= 48.916 g x 4.184 J/g 0C x 1.50C
= 307.0 J
Suppose specific heat capacity of metal is x
Then,
q Metal = (4.967)(x)(-72.7) = -361.1x
The q for the metal is negative because it loses heat and the q for the water is positive because it absorbs heat.
Since the system will reach thermal equilibrium we will make these equal
- q metal = q water
-361.1 x = 307.0
x = 307.0 /361.1
= 0.850J/g0C
For Trial -1
Mass of water = Mass of calorimeter and water - Mass of calorimeter
= 52.852g – 3.869g = 48.983 g
TH2O =Tfinal – T initial
= 22.7 – 21 = 1.7oC
Tmetal =Tfinal – T initial
= 95.2 – 22.7 = 72.5oC
q Water = mc TH2O
c = specific heat capacity of water = 4.184 J/g 0C
q Water = m c t
= 48.983 g x 4.184 J/g 0C x 1.70C
= 348.41 J
Suppose specific heat capacity of metal is x
Then,
q Metal = (5.249)(x)(-72.5) = -380.55x
- q metal = q water
-380.55 x = 348.41
x = 348.41 /380.55
= 0.916J/g0C
Average specific heat = (0.850J/g0C + 0.916J/g0C)/ 2 = 0.883J/g0C
B Heat of solution
For Trial -1
Mass of water = Mass of calorimeter and water - Mass of calorimeter
= 52.446g – 3.869g = 48.577 g
TH2O =Tfinal – T initial
= 16.5 – 21 =- 4.5oC
q Water = mc TH2O
c = specific heat capacity of water = 4.184 J/g 0C
q Water = m c t
= 48.577 g x 4.184 J/g 0C x - 4.50C
= - 914.61 J
H for reaction = - q = - (- 914.61 J) = 914.61 J
For Trial -2
Mass of water = Mass of calorimeter and water - Mass of calorimeter
= 53.046g – 3.869g = 49.177 g
TH2O =Tfinal – T initial
= 16.7 – 21 =- 4.3oC
q Water = mc TH2O
c = specific heat capacity of water = 4.184 J/g 0C
q Water = m c t
= 49.177 g x 4.184 J/g 0C x - 4.30C
= - 884.75 J
H for reaction = - q = - (- 884.75 J) = 884.75 J
Average H for reaction = (914.61 J + 884.75 J)/ 2 = 899.68J
Heat of reaction is endothermic, because H for reaction is positive.
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