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On a phase change graph of temp vs heat, how is the slope inversely proportional to the specific heat?
The equation of the sloped line, according to EK, is q=mcdeltaT. So isn't the slope directly proportional to specific heat since y=mx+b?
Oh, nevermind. In this situation y would be temp and x would be q, so the equation of the line is actually T=q*1/mc so the slope is 1/mc.
Sorry!
The equation of the sloped line, according to EK, is q=mcdeltaT. So isn't the slope directly proportional to specific heat since y=mx+b?
Oh, nevermind. In this situation y would be temp and x would be q, so the equation of the line is actually T=q*1/mc so the slope is 1/mc.
Sorry!
