Okay, I get this.
I caculate the Q for the container and subtract that from the heat of water and take that answer and plug it into find the specific heat of the metal. Right?
Since I know that the amount of energy within the system remains the same. I can apply my value of 21265.75 joules to any equation dealing with the cup, metal or water. So this means, when I place the brick into the water, which is held in the cup, the energy, due to entropy, is being evenly...
Okay, this is what I did
I first found the heat of the water
Q = .925(4180)(31.5-26.0)
Q = 21265.75
Then I plugged Q into the equation for the metal:
21265.75 = .255 C (31.5-95)
Which got me C = 1313 J/kg C
How do I take in account of the container's specific heat? (I'll be back after...
Okay. I figured this out, actually, it takes 1,186,500 Joules (Q = .525KG (2260000 J) (disregarding sig figs), because the added thermal energy for a phase change does not increase the kinetic energy of the particles...right?
Second question:
A container holds 525 grams of water at 100C
How much heat energy must be gained by the water to change it into 100C steam?
Is it 0 Joules due to the fact that it is already at 100C?
I'm a high school student and I have a test soon. I know the basic equation we are using (Q=mCdeltaT), but I having extreme difficultly employing them to get the answer I need, I'm currently stuck on this practice problem:
A 0.925 KG mass of water is place in a container (mass .156 KG) with...