A 0.250-kg coffee cup at 20 degrees C is filled with 0.250 kg of boiling coffee. The cup and the coffee come to thermal equilibrium at 80 degrees C. If no heat is lost the environment, what is the specific heat of the cup material? [Hint: Consider the coffee essentially to be boiling water.]
2007-12-06 22:03:26 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
In geneal
Q=mCp(T2-T1)
Q- heat energy
m mass of the substance
Cp - specific capacity of the substance
T2-T1 - difference in temperature
Q(water)=m(water) Cp(water)(T2-T1)
Q(water)=0.250 x 4.186 (100 -80)=
Q(water)=21J
Q(water)=Q(cup)=m(cup)) Cp(cup)(T1-T3)
Cp(cup)=m(cup)) (T1-T3)/Q(water)
Cp(cup)=0.250 ( 80 - 20)/21=
Cp(cup)=0.714 J/(g K)
Note that Cp's are always using degrees K, however since since we are using Centigrade scale thee difference between two temperature reading will be the same.
2007-12-06 22:13:55 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
Very interesting question... Specific heat, usually designated by the Greek letter gamma, is used in designing many heat devices; one of which is the rocket engines used by NASA. One could assume that the coffee does not make the water in it change in specific heat. It would some, but that would be small. The specific heat of water changes at different temps! But this is also very minor. The cup grabbed 20 degrees from the coffee. The rise in temp of the cup is 60 degrees and the lost in temp of the boiling coffee is 20 degrees. That means it must have a specific heat value of 1/3 of the coffee, considered as boiling water... Earl
2007-12-06 22:50:58 · answer #2 · answered by ? 6 · 0⤊ 0⤋
yeah,,,whatever he said =D
2007-12-06 22:24:08 · answer #3 · answered by kluivert(17) 4 · 1⤊ 0⤋