A 24.0-g sample of an unknown metal is heated to 94.0°C and is placed in a perfectly insulated container along with 196.g of water at an initial temperature of 20.15°C. After a short time the temperature of both the metal and water become equal at 23.80°C.
The specific Heat Capacity of water is 4.18 J/g/K in this temperature range.
What is the specific heat capacity of the metal?
Can anyone help me with this question?
Please provide me with solution
Thanks alot:))
2007-06-16 15:07:15 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
The formula is q = m Cp delta T
First calculate the heat gained by the water. You have the mass, Cp and delta T
now set this heat equal to the energy lost by the metal. You have the mass and delta T. Now you can solve for the Cp
2007-06-16 15:09:58 · answer #1 · answered by reb1240 7 · 0⤊ 0⤋
THat answer is Cp = .239 J / g ok this suggests which you fairly elect 4.1855 J of capacity to upward thrust a million °C of one gram of water . different one is Cp = 4.1855 kJ / kg : which potential you pick 4.1855 KJ to upward thrust a million °C of one Kilogram of water. This answer is powerful because of the fact i'm a chemistry instructor.
2016-12-08 11:16:52 · answer #2 · answered by Erika 4 · 0⤊ 0⤋
oh man
we're doing the same thing... and a government exam next
week and i dont get anything :S
good luck
2007-06-16 15:15:46 · answer #3 · answered by itssoeasy 6 · 0⤊ 0⤋
m(metal)*SH(Metal)* delta T(metal) = -m(water)*SH(water)*delta T(water)
SH(metal) = -196*4.184*3.65/(24.0*-70.2) = 1.78 J/g K
2007-06-16 15:18:17 · answer #4 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋