A 14.3 g sample of an unknown metal is heated to 94.0 oC and is placed in a coffee cup calorimeter containing 122 g of water at a temperature of 20.45 oC. After the metal cools, the final temperature of the metal and water is 23.35 oC. Calculate the specific heat capacity (in J/goC) of the metal, assuming that no heat escapes to the surroundings. (The specific heat capacity of water = 4.184 J/goC)
2007-04-06 15:45:56 · 2 answers · asked by lola c 2 in Science & Mathematics ➔ Chemistry
I answered a similar question like this earlier...
Umm... let's see... You calculate the energy of water first.
The formula to calculate energy is q= (m)(cp)(delta T)
m= mass, cp= specific heat, delta t = change in temp.
change in temp = final temp. - initial temp.
q= (122) (4.184) (23.35- 20.45)
q=1480. 2992 J
Since water gained energy of +1480. 2992 J, then that means the unknown metal lost -1480.2992 J according to law of conservation of energy.
So the equation to calculate specific heat is q/ (m)(delta T).
cp= -1480.2992 J/ (14.3g) (23.35-94.0)
cp= 1.4652.... J/g degrees Celcius..
Umm... I'm too lazy to figure out the number significant figures in your answer, so I'll leave that for you to figure out. :P
2007-04-06 17:07:40 · answer #1 · answered by Trish 2 · 0⤊ 0⤋
(14.3 g)(94.0°C - 23.35°C)u = (122 g)(4.186 J/g°C)(23.35°C - 20.45°C)
u = (122)(4.186 J/g°C)(2.9)/(14.3)(70.65)
u â 1.4659 J/g°C
2007-04-06 23:20:58 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋