Respuesta :
Answer : The final temperature of the mixture is [tex]61.4^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]
And as we know that,
Mass = Density × Volume
Thus, the formula becomes,
[tex](\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = [tex]c_2[/tex] = specific heat of water = same
[tex]m_1[/tex] = [tex]m_2[/tex] = mass of water = same
[tex]\rho_1[/tex] = [tex]\rho_2[/tex] = density of water = 1.0 g/mL
[tex]V_1[/tex] = volume of water at [tex]80.0^oC[/tex] = [tex]370cm^3=370mL[/tex]
[tex]V_2[/tex] = volume of water at [tex]4^oC[/tex] = [tex]120cm^3=120mL[/tex]
[tex]T_f[/tex] = final temperature of mixture = ?
[tex]T_1[/tex] = initial temperature of water = [tex]80.0^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]4^oC[/tex]
Now put all the given values in the above formula, we get:
[tex](\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)[/tex]
[tex](1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC[/tex]
[tex]T_f=61.4^oC[/tex]
Therefore, the final temperature of the mixture is [tex]61.4^oC[/tex]
