rach asked:


Describe an experiment using only hot and cold water that would enable you to verify that the heat capacity of a coffeecup calorimter is 1.0 * 10^1 J/C.

Add videos-audio-webcams

Chad asked:


a 51.6ml dillute solution of acid at 23.85C is mixed with 48.5ml of a dilute solution of base, also at 23.85C, in a coffeecup calorimeter. After the reaction occurs, the temperature of the resulting mixture is 27.25C the density of the final solution is 1.03 g/ml. calculate the amount of heat evolved. assume the specific heat of the solution is 4.184 J/g*C. the heat capacity of the calormeter is 23.9 J/C.

Best Web Design Software

js142004 asked:


I am about to make a planning experiment on the topic: Measure energy content in various types of food? Any idea how to do this measure?

Best Hosting

erika86 asked:


Time=0 Initial temperature was 24.3 Celsius

Time(s) t(C)
30 -> 38.6 Celsius
60 -> 39.7 C
90 -> 39.5 C
120 -> 39.4 C
150 -> 39.3 C
180 -> 39.2 C
210 -> 39.1 C
240 -> 39.0 C

Is this an exothermic or an endorthermic reaction? Why?
Why does the temperature increase, reach a maximum, and then decrease?

Please help me!!!

Get support building a website

Julie R asked:


What is in it and where can I find that info online?

Best Web Design Software

Kitty asked:


Im doing a lab based on a coffeecup calorimeter.
I have my data, which i plotted on a graph
i put 80deg.C water in a beaker of 100 mL room temperature water (26degC) and the temperature shot up to 48 and then it stabilized around 47 degrees. Now, the question says: “After several measurements, the points should form a straight line as the temperature slowly decreases, showing the loss of heat to the calorimeter assembly. draw a straight line through these points back to time zero to find (extrapolate) the theoretical T(sub0) temperature correcting for equilibrium of the equipment
What i did was take the average of 47 and 26 (starting temperature of water before hot water was put in), which gave me 36.5
But then, when I go to calculate the heat lost, Q(sub w)=mcdeltaT
m is 100 g because i used 100 mL, c is specific heat of water, 4.18 J/(g*C) and…then the delta T is where I get lost
It says that delta T is T(sub 0)-Taverage
T average is avg. of 80C and room temp (26C)=53
so then, the T(sub 0) i calculated was 36.5 (the average i took, whcih is prob wrong)
36.5-53 (Taverage)=-16.5C.
Its not supposed to be negative?!?

what did i do wrong????

The whole T(sub 0) is throwing me off.
help, please! any at all is GREATLY appreciated!!!
It is throwing off my calculations because i have to use T(sub 0) in the next parts of the question..

Add videos-audio-webcams

Matt H asked:


Our company is working on a few buildings installing exterior clad granite. One of our stones has a higher iron content than the other yet it has begun to rust, while the other has not at all. I have also seen this just in different parts of a quarry. What else would lead to this. I have a BS in Biology so I understand chemical reactions and have a basis for this, but cannot answer it fully.

High Rated Web Design Software