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| by Dejan Tojcic
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| ==Specific Heat Capacity ==
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| Specific Heat Capacity is simply a physical quantity that represents the ratio of the amount of heat taken or added to substance or object which results in a temperature change. The formal definition of Specific Heat is the amount of heat required to raise the temperature of 1 gram of a substance 1°C. The Standard Unit(SI) of this quantity is, joule per celsius per kilogram or or <math>\mathrm{\tfrac{J}{°C*Kg}}</math>. Different objects/substances have different specific heat capacities because every object has a varying mass, molecular structure, and numbers of particles per unit mass specific, and since specific heat capacity is reliant on mass, every different object has a different specific heat capacity.
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| [[File:specificheatmetals.jpg]]
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| ===A Mathematical Model===
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| In order to find the Specific Heat Capacity of a substance, we use the equation:<math> \Delta E_{\mathrm{thermal}} = C * M * \Delta T </math>, and rearrange it to get <math> C= \Delta E_{\mathrm{thermal}} /( M * \Delta T) .</math> where ''C'' is the Specific Heat Capacity with units of joules per celsius per kilogram or <math>\mathrm{\tfrac{J}{°C*Kg}}</math>, ''M'' is the mass measured in kilograms or <math>\mathrm{kg}</math>, ''<math> \Delta E_{\mathrm{thermal}} </math>'' represents the change in thermal energy measured by joules or <math>\mathrm{J}</math>, and ''<math> \Delta T </math>'' represents change in temperature with units celsius or <math>\mathrm{°C.}</math>
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| [[File:picture2.jpg]]
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| ===A Computational Model===
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| How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]
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| =Simple Example=
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| '''''Question'''''
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| It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g·°C?
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| [[File:Zvezda33.jpg]]
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| '''''Solution'''''
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| This problem is very simplistic in nature as we are simply need to plug all of the values into the equation.
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| Remembering the equation
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| [[File:Zvezda5.jpg]]
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| All we have to do is rearrange it and solve for the specific heat capacity
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| [[File:zvezda6.jpg]]
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| Knowing that ''<math> \ Q </math>'' or ''<math> \Delta E_{\mathrm{thermal}} </math>'' is equal to the change in thermal energy, we can plug in ''487.5 Joules'' into the equation, so now we have
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| <math> C= 487.5 joules /( M * \Delta T) .</math>
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| Next, looking at the question it looks like we are given the mass or <math>\mathrm{M}</math> which is ''25 grams'' or . And can now plug that into the equation to get
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| <math> C= 487.5 joules /(25g * \Delta T) .</math>
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| Lastly, when we read the problem we see that we are given ''<math> \Delta T </math>'' which in this case would be <math> (75 °C-25 °C) </math> or <math> 50 °C </math>. We can finally solve for the value of <math>\mathrm{C}</math> which is
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| <math> C= 487.5 joules /( 25g * 50 °C) .</math>
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| and we finally conclude that
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| <math> C= .39{\tfrac{J}{g*°C}}</math> or the specific heat of copper is .39 J/(g*°C).
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| =Intermediate example=
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| '''''Question''''' What is the final temperature when 625 grams of water at 75.0° C loses 7.96 x 10^4 J?
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| [[File:zvezda7.jpg]]
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| ----
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| '''Solution'''
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| =Hard Example=
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| =Connectedness=
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| #How is this topic connected to something that you are interested in?
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| #How is it connected to your major?
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| #Is there an interesting industrial application?
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| ==History==
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| Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
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| == See also ==
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| Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?
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| ===Further reading===
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| Books, Articles or other print media on this topic
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| ===External links===
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| Internet resources on this topic
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| ==References==
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| This section contains the the references you used while writing this page
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| [[Category:Which Category did you place this in?]]
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