Answer: The final temperature of the mixture will be [tex]28.5^0C[/tex]
Explanation:
[tex]heat_{absorbed}=heat_{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_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 mercury = 425 g
[tex]m_2[/tex] = mass of water = 145 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of mercury = [tex]85.0^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]23.0^oC[/tex]
[tex]c_1[/tex] = specific heat of mercury = [tex]0.140J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water= [tex]4.18J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]-425\times 0.140\times (T_{final}-85.0)=[145\times 4.184\times (T_{final}-23.0)][/tex]
[tex]T_{final}=28.5^0C[/tex]
Therefore, the final temperature of the mixture will be [tex]28.5^0C[/tex]
