Answer : The specific heat capacity of the mineral is [tex]0.131J/g^oC[/tex]
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of unknown mineral = ?
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of unknown mineral = 307 g
[tex]m_2[/tex] = mass of water = 72.4 g
[tex]T_f[/tex] = final temperature of calorimeter = [tex]32.4^oC[/tex]
[tex]T_1[/tex] = initial temperature of unknown mineral = [tex]98.7^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]23.6^oC[/tex]
Now put all the given values in the above formula, we get
[tex]307g\times c_1\times (32.4-98.7)^oC=-72.4g\times 4.18J/g^oC\times (32.4-23.6)^oC[/tex]
[tex]c_1=0.131J/g^oC[/tex]
Therefore, the specific heat capacity of the mineral is [tex]0.131J/g^oC[/tex]