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Three liquids are at temperatures of 13 ◦C, 19◦C, and 40◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 15◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 36.9 ◦C. Find the equilibrium temperature when equal masses of the first and third are mixed. Answer in units of ◦C.

Respuesta :

Answer:

2 C1 = 4 C2         heat loss = heat gain

C1 = 2 C2

17.9 C2 = 3.1 C3      mixing second and third

17.9 / 2 C1 = 3.1 C3         C1 for C2

17.9 C1 = 6.2 C3

C3 = 2.89 C1

C1 (T - 13) = C3 (40 - T)       heat loss = heat gain

T - 13 = 2.89 (40 - T) = 115.5 - 2.89 T

3.89 T = 128.5

T = 33 deg C

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