Answer: The final temperature will be [tex]23^oC[/tex]
Explanation:
Calculating the heat released or absorbed for the process:
[tex]q=m\times C\times (T_2-T_1)[/tex]
In a system, the total amount of heat released is equal to the total amount of heat absorbed.
[tex]q_1=-q_2[/tex]
OR
[tex]m_1\times C_1\times (T_f-T_1)=-m_2\times C_2\times (T_f-T_2)[/tex] ......(1)
where,
[tex]C_1[/tex] = specific heat of aluminium = [tex]0.21 Cal/g^oC[/tex]
[tex]C_2[/tex] = heat capacity of water = [tex]1Cal/g^oC[/tex]
[tex]m_1[/tex] = mass of aluminium = 30. g
[tex]m_2[/tex] = mass of water = 50. g
[tex]T_f[/tex] = final temperature of the system = ?
[tex]T_1[/tex] = initial temperature of aluminium = [tex]40.^oC[/tex]
[tex]T_2[/tex] = initial temperature of the water = [tex]21.^oC[/tex]
Putting values in equation 1, we get:
[tex]30\times 0.21\times (T_f-40)=-50\times 1\times (T_f-21)\\\\56.3T_f=1302\\\\T_f=\frac{1302}{56.3}=23.13^oC=23^oC[/tex]
Hence, the final temperature will be [tex]23^oC[/tex]
