Question

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 1

Answer : The final temperature of the mixture is
`22.7^oC`

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

`(\rho_1* V_1)* c_1* (T_f-T_1)=-(\rho_2* V_2)* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of ethanol =
`2.3J/g^oC`

`c_2` = specific heat of water =
`4.18J/g^oC`

`m_1` = mass of ethanol

`m_2` = mass of water

`\rho_1` = density of ethanol = 0.789 g/mL

`\rho_2` = density of water = 1.0 g/mL

`V_1` = volume of ethanol = 45.0 mL

`V_2` = volume of water = 45.0 mL

`T_f` = final temperature of mixture = ?

`T_1` = initial temperature of ethanol =
`9.0^oC`

`T_2` = initial temperature of water =
`28.6^oC`

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

`(0.789g/mL* 45.0mL)* (2.3J/g^oC)* (T_f-9.0)^oC=-(1.0g/mL* 45.0mL)* 4.18J/g^oC* (T_f-28.6)^oC`

`T_f=22.7^oC`

Therefore, the final temperature of the mixture is
`22.7^oC`