I have a diffficult calorimetry problem I could use assistance with?
Initial temperature of water in calorimeter = 22.0 oC
Mass of unknown metal = 45.0 g
Initial temperature of hot metal = 100.0 oC
Final temperature of water after you place hot metal in calorimeter filled with water = 28.0 oC
Based on the above data, what is the Cp of the metal?
(Cp H2O = 4.184 J/g oC )
Answer:
(28.0 - 22.0) (90.0 g) (4.184 J/g) = 2259.360 J
(100.0 C - 28.0 C) (45.0 g) (Cp Metal) = 2259.360 J
Cp = .697 J/g/degree
Keep in mind that the heat lost by the hot metal is the same as the heat gained by the water (it looks like they're assuming the calorimeter isn't gaining any heat, although technically it would)
In mathematical form this would look like:
heat change of metal = -heat change of water
mass metal x Cp metal x (Tf metal - Ti metal) = -mass water x Cp water x (Tf water - Ti water)
Fill in the known values:
45 x Cp metal x (28 - 100) = -90 x 4.184 x (28 - 22)
Solve for Cp metal:
Cp metal = 0.70 J/g C
heat lost by metal = 45*Cp*72
heat gained by water = 90*4.184*6
According to law of calorimetry,
Heat gained= Heat lost
So,
45*C*72 = 90*4.184*6
Calculating value of C we get C= 0.69733 J/g/ oC
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: