Answer : The final temperature of the system is, [tex]27.9^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 gold = [tex]0.129J/g.^oC[/tex]
Note : Molar heat capacity of gold = [tex]25.41J/mol.^oC[/tex]
Heat capacity of gold = [tex]\frac{25.41J/mol.^oC}{197g/mol}=0.129J/g.^oC[/tex]
[tex]c_2[/tex] = heat capacity of water = [tex]4.18J/g.^oC[/tex]
[tex]m_1[/tex] = mass of gold = 13.00 g
[tex]m_2[/tex] = mass of water = 13.00 g
[tex]T_f[/tex] = final temperature of system = ?
[tex]T_1[/tex] = initial temperature of gold = [tex]90.0^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]26.00^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]13.00g\times 0.129J/g.^oC\times (T_f-90.0)^oC=-13.00g\times 4.18J/g.^oC\times (T_f-26.00)^oC[/tex]
[tex]T_f=27.9^oC[/tex]
Therefore, the final temperature of the system is, [tex]27.9^oC[/tex]
