Question

Three liquids are at temperatures of 9˚C,

20°C, and 38°C, respectively. Equal masses of
the first two liquids are mixed, and the equi-
librium temperature is 16°C. Equal masses of
the second and third are then mixed, and the
equilibrium temperature is 32.1°C.
Find the equilibrium temperature when
equal masses of the first and third are mixed.
Answer in units of °C.

Answer 1

Let the specific heat of the first liquid be c1, the specific heat of the second liquid be c2, and the specific heat of the third liquid be c3. Let the mass of each liquid be m.

When the first two liquids are mixed, the heat lost by the warmer liquid is equal to the heat gained by the cooler liquid. Using the formula for heat transfer:

(m * c2 * (16°C - 20°C)) = (m * c1 * (20°C - 9°C))

Simplifying this equation gives:

-4m c2 = 11m c1

Dividing both sides by m and rearranging:

c1/c2 = -4/11

When the second and third liquids are mixed, we can use the same formula:

(m * c2 * (32.1°C - 20°C)) = (m * c3 * (38°C - 32.1°C))

Simplifying this equation gives:

12.1m c2 = 5.9m c3

Dividing both sides by m and using the ratio we found earlier:

c3/c2 = 12.1/5.9 * c1/c2 = (12.1/5.9) * (-4/11) = -1.6

Now we can use the same formula again to find the equilibrium temperature when the first and third liquids are mixed:

(m * c1 * (T - 9°C)) = (m * c3 * (38°C - T))

Simplifying this equation gives:

29m c1 + m c3 T = 342m c3

Dividing both sides by m and using the ratio we found earlier:

29c1 - 1.6c3 = 342c3

Solving for T:

T = (29c1 - 342c3) / 1.6c3

Substituting the specific heat ratios we found earlier:

T = (29 - 342c3/c1) / 1.6

Substituting c3/c1 = -11/4 from the first ratio we found:

T = (29 + 342/4) / 1.6 = 61.44°C

Therefore, the equilibrium temperature when equal masses of the first and third liquids are mixed is 61.44°C.