To find the specific heat capacity of a certain metal, a student places a block of the metal that weighs 27 grams and has an initial temperature of 97 deg C, into 52 grams of water with a temperature of 20 deg C. The final temperature was measured to be 27 deg C. What is the heat capacity of the metal?

Respuesta :

Answer:

The heat capacity of the metal is: [tex]0.8043\frac{kJ}{kgK}[/tex]

Explanation:

Consider the energy balance; it says that the energy in the initial state is equal to the energy in the final state for the system, where the system is the water and the metal. M and W subscripts mean metal and water respectively, and 1 and 2 superscripts means the initial an the final state respectively.

The energy balance is:

[tex]{U_{M}}^1+{U_W}^1={U_{M}}^2+{U_W}^2\\{U_{M}}^1-{U_{M}}^2={U_{W}}^2-{U_{W}}^1\\m_M*Cv_M*({T_{M}}^1-{T_{M}}^2)=m_W*Cv_W*({T_{W}}^2-{T_{W}}^1)[/tex]

The difference between Cp and Cv is neglective for solids and liquids, so it is possible to change Cv by Cp:

[tex]m_M*Cp_M*({T_{M}}^1-{T_{M}}^2)=m_W*Cp_W*({T_{W}}^2-{T_{W}}^1)[/tex]

And the only unknown from the equation is [tex]Cp_M[/tex], so:

[tex]Cp_M=\frac{Cp_W*({T_W}^2-{T_W}^1)*m_W}{({TM}^1-{T_M}^2)*m_M}\\Cp_M=\frac{4.176\frac{kJ}{kgK}*(27-20)C*0.052kg}{(97-27)C*0.027kg}\\ Cp_M=0.8043\frac{kJ}{kgK}[/tex]

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