Three liquids are at temperatures of 6 ◦C, 19◦C, and 34◦C, respectively.?
Equal masses
of the first two liquids are mixed, and the
equilibrium temperature is 9◦C. Equal masses
of the second and third are then mixed, and
the equilibrium temperature is 31.1
◦C. Find the equilibrium temperature when
equal masses of the first and third are mixed.
Answer in units of ◦C
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Verified answer
first mix:
m * c1 * (9 - 6) = m * c2 * (19 - 9) → mass m cancels
c2 = c1 * 3 / 10 = 0.3*c1
second mix:
(m * c1 + m * c2) * (31.1 - 9) = m * c3 * (34 - 31.1) → m cancels
(c1 + 0.3c1) * 22.1 = c3 * 2.9
c3 = 1.3c1 * 22.1 / 2.9 = 9.9069c1
third mix:
m * c1 * (T - 6) = m * 9.9069c1 * (34 - T) → mass m and c1 cancels
and the rest solves to
T = 31.4ºC
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I agree with the Best Answer. However, for the second mix, there should be nothing from the first mix's answer except for 0.3c1. Essentially, the left side of the second mix equation should look like: m *0.3c1 * (31.1-19). Then, you can properly solve the equation.