Respuesta :
Answer : The final temperature of the mixture is [tex]45^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\times (T_f-T_1)=-m_2\times c\times (T_f-T_2)[/tex]
And as we know that,
Mass = Density × Volume
Thus, the formula becomes,
[tex](\rho\times V_1)\times c\times (T_f-T_1)=-(\rho\times V_2)\times c\times (T_f-T_2)[/tex]
where,
c = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]\rho[/tex] = density of water = 1.00 g/mL
[tex]V_1[/tex] = volume of water at [tex]25.00^oC[/tex] = 300.0 mL
[tex]V_2[/tex] = volume of water at [tex]95.00^oC[/tex] = 120.0 mL
[tex]T_f[/tex] = final temperature of mixture = ?
[tex]T_1[/tex] = initial temperature of water = [tex]25.00^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]95.00^oC[/tex]
Now put all the given values in the above formula, we get
[tex](1.00g/mL\times 300.0mL)\times (4.18J/g^oC)\times (T_f-25.00)^oC=-(1.0g/mL\times 120.0mL)\times 4.18J/g^oC\times (T_f-95.00)^oC[/tex]
[tex]T_f=45^oC[/tex]
Therefore, the final temperature of the mixture is [tex]45^oC[/tex]
