Answer : The specific heat of metal is, [tex]0.928J/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 metal = ?
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of metal = 50 g
[tex]m_2[/tex] = mass of water = 100 g
[tex]T_f[/tex] = final temperature = [tex]30^oC[/tex]
[tex]T_1[/tex] = initial temperature of metal = [tex]120^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]20^oC[/tex]
Now put all the given values in the above formula, we get
[tex]50g\times c_1\times (30-120)^oC=-100g\times 4.18J/g^oC\times (30-20)^oC[/tex]
[tex]c_1=0.928J/g^oC[/tex]
Therefore, the specific heat of metal is, [tex]0.928J/g^oC[/tex]
