A 20 g piece of aluminum at 0C is dropped into a beaker of water. The temp of the water drops from 90C to 75C. What quantity of heat energy did the peice aluminum abosrb? The specific heat of aluminum is .902 J/gC.
to do this problem do you do this:
q=mst
q=(20g)(.902)(75-90)
?? Thanks.
2007-12-30 06:23:55 · 3 answers · asked by senseless.student 1 in Science & Mathematics ➔ Chemistry
Isn't the answer supposed to be positive since the reaction absorbs heat? endothermic? idk
2007-12-30 06:29:54 · update #1
The aluminum block rises in temperature from 0C to 75C, so the temperature increase is 75C.
Specific heat capacity = 0.902 J/gC - i.e. 0.902 Joules are required for each g raised by 1C
We are raising 20g by 75C = 1500gC
Hence energy absorbed = 0.902*1500= 1353 J
2007-12-30 06:31:54 · answer #1 · answered by yodellingbear 3 · 0⤊ 0⤋
no suitable gas equation isn't used at okay here. the thermodynamics equation is the suitable technique as a results of fact the questioner stated. the place deltaE is the exchange in int skill , q is the warmth absorbed by utilizing the gas and w is the artwork completed on the gas. the question mentions that the gadget will enhance in quantity so the gas does artwork on the encompassing(artwork is accomplished by utilizing the gas no longer on the gas). this constitutes a average loss in IE in certainty. this loss is comparable to -(p × deltaV), the place p is rigidity in pascals, V is quantity in m³. so the thermodynamic equation sounds like this -102.5=+fifty two.5-p(deltaV) on the grounds which you understand p (purely convert atm to pascal), resolve the equation for deltaV, and artwork the respond into the outstanding quantity of 0.058m³ to get the intial quantity. physics is relaxing isn't it?
2016-10-09 21:40:24 · answer #2 · answered by ? 4 · 0⤊ 0⤋
That's right.
2007-12-30 06:28:25 · answer #3 · answered by steve_geo1 7 · 0⤊ 0⤋