Answer: [tex]0.40J/g^0C[/tex]
Explanation:
[tex]heat_{released}=heat_{absorbed}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)][/tex] .................(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of metal = 100.0 g
[tex]m_2[/tex] = mass of water = 100.0 g
[tex]T_{final}[/tex] = final temperature = [tex]80^0C[/tex]
[tex]T_1[/tex] = temperature of metal = [tex]600^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]30^oC[/tex]
[tex]c_1[/tex] = specific heat of metal = ?
[tex]c_2[/tex] = specific heat of water= [tex]4.2J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]-[100.0\times c_1\times (80-600)]=[100.0\times 4.2\times (80-30)][/tex]
[tex]c_1=0.404J/g^0C[/tex]
Therefore, the specific heat capacity of metal is [tex]c_1=0.40J/g^0C[/tex]
