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Calorimeter and heat, units question

  • Thread starter fluidistic
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  • #1
fluidistic
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Homework Statement


I have a lab report due next week and I'm almost done with it. However I'm totally stuck on the last part where it is wrote in a description of the experience : [tex]c=\frac{(m+M_e)(T_f'-T'_i)}{m(T_c-T_f)}[/tex] where [tex]c[/tex] is the specific heat of a metal. If I work out the units in the formula, I end up with [tex]c[/tex] has no units! It must have units of [tex]\frac{cal}{g \cdot K}}[/tex]! But I don't know how to make this appear!
The only thing I imagine that can save me is that the [tex]c[/tex] above is equivalent to one with units, but I don't know how to justify this. Clearly it makes no sense at all to me.
I'd be very grateful if you could help me. (I'd be so glad to finish the report as soon as possible so that I can study vector analysis...)
 

Answers and Replies

  • #2
mgb_phys
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I think you need a C of the calorimeter metal on the top.
 
  • #3
fluidistic
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I think you need a C of the calorimeter metal on the top.
Sorry I don't understand what you mean.
This formula also appear in some websites, like http://www.sc.ehu.es/sbweb/fisica/estadistica/otros/calorimetro/calorimetro.htm (In Spanish. Where it says
El calor específico desconocido del será por tanto...
)
In my formula [tex]M_e[/tex] is the mass of water to which the calorimeter is equivalent to.
 

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