posted by April .
Can someone please help me with the following?
How do I find the specific heat of a metal when the only information I have is a temperature change ( of -73 degrees C) and a mass (of 22.53 grams)?
And what is the forumla for specific heat?
Thanks in advance!
Hmmm...seems like you would need one more thing...
The equation is:
Q = M c (deltaT)
where Q is the amount of heat put in, M is the mass, and deltaT is the change in temperature. Do they tell you Q??
Does c stand for the specific heat because that is what I am finding? And I don't know the Q, which is the Joules or calories, right? Or is the c the calories?
Yes, the Q is the joules...the amount of heat put in. c is the specific heat.
Okay, well, I have a metal and I'm supposed to identify it by figuring out its specific heat. I did a lab and all it requires of me is to find the mass and temperature change. How do I go about finding the specific heat when I don't know the Joules?
How did you do the experiment? Did you add the hot metal to water, and then record the temperature change of the water?
Yeah. Umm, I found the formula! I have everything for it other than the mass of the water. I know the volume of the water, but not the mass. Is it possible to find the mass of water without its density?
yes, the density of water is a "given." it is 1 gm/ml
Okay, thank you! I think I can figure it out from there. :)
You cant. You must have something else, normally, the heat absorbed.
In High School, you often do this in a water bath (known mass of water, known specific heat, and a measured temperature change). If that is what you did, then the water lost heat (negative heat gain) as the metal absorbed heat.
Heatgainedwater + Heatgainedmetal=0
(one will turn out negative to add to zero)
mw*cw*(Tfinal-tinitialwater) + mm*cm*(Tfinal - Tinitialmetal)=0
you can solve for cm from this.
Thanks, but that's not what we did.
OK, without some measure of the heat absorbed to which increased the temperature of the metal, the specific heat cannot be determined.
It can. I found the formula in my chemistry lab packet and DanH helped me figure out how.