Question

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?

Answer 1

The heat capacity of the metal is:
`0.8043(kJ)/(kgK)`

Step-by-step 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:

`{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)`

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

`m_M*Cp_M*({T_(M)}^1-{T_(M)}^2)=m_W*Cp_W*({T_(W)}^2-{T_(W)}^1)`

And the only unknown from the equation is
`Cp_M`, so:

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