Answer : The final temperature of the metal block is, [tex]25^oC[/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 aluminum = 55 g
[tex]m_2[/tex] = mass of water = 0.48 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of aluminum = [tex]25^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]25^oC[/tex]
[tex]c_1[/tex] = specific heat of aluminum = [tex]0.900J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of water= [tex]4.184J/g^oC[/tex]
Now put all the given values in equation (1), we get
[tex]55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC][/tex]
[tex]T_{final}=25^oC[/tex]
Thus, the final temperature of the metal block is, [tex]25^oC[/tex]
