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wow, this one is thermodynamics right?
its actually physics but it's considered to be in the math category so i need a lot of help with this.
The heat associated with a certain mass changing temperature is
the energy change in the water has to equal the energy change in the metal
you set the two equal to each other and solve for c. That equation will give units of Joule/(gram Celsius)
so would i be solving the equation that you suggested i'm confused
Have you learned the heat/energy required to change the temperature of a substance?
no i don't quite understand how to solve problems like these.
ok we have the equation Q=mc(dT) where Q is heat, m is mass of whatever is changing temperature, dT is the change in Temperature.
the magnitude of the heat has to be the same for both the metal and the water
so you plug the known values into the heat equation for water and metal and set them equal to each other to find the specific heat you are looking for.
The magnitude of the heat is the same in both due to conservation of energy
so basically what would be the numbers in the equation to plug into the problem if is the same.
Your final equation would look like: 140*c*(70-10)=220*4.186*(10-0) 140*60*c=220*10*4.186
and you want to find c in J/gC
so how would i find c
140*60*c=220*10*4.186 8400*c=9209.2 c=9209.2/(8400) c=1.1 J/gC
1.0963 would be c
so what units should i use for this problem?
that c is J/(gC) or Joules per gram degree Celsius
the energy in Joules it takes to change the temperature of 1 gram of metal 1 degree celsius
so would this be the same as 1.0963 cal/g degree Celsius ?
no 1 cal= 4.186 Joules
if you want calories divide your answer by 4.186
i don't think my answer is right
i have 0.2619
yep I think that's good
why dont you think it is right
i don't know because i think that my units wont match up with the right answer that i just put up there
im not sure what you mean
like should it say 0.2619 cal/g degrees Celsius, does that sound right?
oh yeah those are the units for specific heat
thank you so much I got it right thanks
you're welcome :)