Answer:c=0.213 J/g/K
Explanation:
Given
sample mass[tex](m_s)=120 gm[/tex]
initial temperature of mineral[tex](T_{im})=92.6[/tex]
mass of water[tex](m_w)=82.9[/tex]
Water initial temperature[tex](T_{iw})=20[/tex]
Heat capacity of calorimeter =15.3 J/k
Final Temperature is 24.8
let c be the specific heat of mineral
Heat released by mineral sample=heat absorbed by calorimeter and water
[tex]120\times c\times (92.6-24.8)=82.9\times 4.184\times (24.8-20)+15.3\times (24.8-20)[/tex]
[tex]c\times 67.8=14.486[/tex]
[tex]c=0.213 J/g/K[/tex]