Respuesta :
Answer : The final temperature of the aluminum and ethylene glycol is [tex]56.2^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 aluminum = [tex]0.900J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of ethylene glycol = [tex]2.36J/g^oC[/tex]
[tex]m_1[/tex] = mass of aluminum = 13.60 g
[tex]m_2[/tex] = mass of ethylene glycol = 23.27 g
[tex]T_f[/tex] = final temperature of aluminum and ethylene glycol = ?
[tex]T_1[/tex] = initial temperature of aluminium = [tex]13.91^oC[/tex]
[tex]T_2[/tex] = initial temperature of ethylene glycol = [tex]65.66^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]13.60g\times 0.900J/g^oC\times (T_f-13.91)^oC=-23.27g\times 2.36J/g^oC\times (T_f-65.66)^oC[/tex]
[tex]T_f=56.2^oC[/tex]
Therefore, the final temperature of the aluminum and ethylene glycol is [tex]56.2^oC[/tex]
