Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.
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_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)][/tex] .................(1)
where,
q = heat absorbed or released
[tex]m_A[/tex] = mass of A = 2x
[tex]m_B[/tex] = mass of B = x
[tex]T_{final}[/tex] = final temperature = z
[tex]T_A[/tex] = temperature of A
[tex]T_2[/tex] = temperature of B
[tex]c_A[/tex] = specific heat capacity of A = y
[tex]c_B[/tex] = specific heat capacity of B = 2y
Now put all the given values in equation (1), we get
[tex]2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)][/tex]
[tex]2z=T_B+T_A[/tex]
[tex]z=\frac{T_B+T_A}{2}[/tex]
Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.