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If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.6 C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?



I tried many times in trying to get the answer, but I keep getting it wrong. I appreciate who answers this writes it step by step. Thank you.

Answer :

Answer : The final temperature of the mixture is [tex]22.7^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] = specific heat of ethanol = [tex]2.3J/g^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of ethanol

[tex]m_2[/tex] = mass of water

[tex]\rho_1[/tex] = density of ethanol = 0.789 g/mL

[tex]\rho_2[/tex] = density of water = 1.0 g/mL

[tex]V_1[/tex] = volume of ethanol = 45.0 mL

[tex]V_2[/tex] = volume of water = 45.0 mL

[tex]T_f[/tex] = final temperature of mixture = ?

[tex]T_1[/tex] = initial temperature of ethanol = [tex]9.0^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]28.6^oC[/tex]

Now put all the given values in the above formula, we get

[tex](0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC[/tex]

[tex]T_f=22.7^oC[/tex]

Therefore, the final temperature of the mixture is [tex]22.7^oC[/tex]

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