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Pie
3 years ago
13

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.
Chemistry
1 answer:
Klio2033 [76]3 years ago
4 0

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\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\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)

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\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

T_f=22.7^oC

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

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