Answer :
Answer: The specific heat capacity of metal is [tex]1.31J/g^0C[/tex]
Explanation:
[tex]Q_{absorbed}=Q_{released}[/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\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of metal = 35 g
[tex]m_2[/tex] = mass of water = 220 g
[tex]T_{final}[/tex] = final temperature = [tex]30^0C[/tex]
[tex]T_1[/tex] = temperature of metal = [tex]130^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]25^oC[/tex]
[tex]c_1[/tex] = specific heat of metal = ?
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]
[tex]35\times c_1\times (30-130)^0C=-[220g\times 4.184\times (30-25)][/tex]
[tex]c_1=1.31J/g^0C[/tex]
Therefore, the specific heat capacity of metal is [tex]1.31J/g^0C[/tex]