Respuesta :
Answer:
16.1°C
Explanation:
When the first two are mixed:
m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0
C₁ (T₁ − T) + C₂ (T₂ − T) = 0
C₁ (14 − 16) + C₂ (25 − 16) = 0
-2 C₁ + 9 C₂ = 0
C₁ = 4.5 C₂
When the second and third are mixed:
m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0
C₂ (T₂ − T) + C₃ (T₃ − T) = 0
C₂ (25 − 27.9) + C₃ (33 − 27.9) = 0
-2.9 C₂ + 5.1 C₃ = 0
C₂ = 1.76 C₃
Substituting:
C₁ = 4.5 (1.76 C₃)
C₁ = 7.91 C₃
When the first and third are mixed:
m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0
C₁ (T₁ − T) + C₃ (T₃ − T) = 0
(7.91 C₃) (14 − T) + C₃ (33 − T) = 0
(7.91) (14 − T) + 33 − T = 0
111 − 7.91T + 33 − T = 0
8.91T = 144
T = 16.1°C
In the thermal equilibrium, the net heat transfer is said to be zero in between the bodies. The equilibrium temperature when equal masses of the first and third are mixed will be 16.1° C.
What is thermal equilibrium?
Thermal equilibrium is easily explained by the zeroth law of thermodynamics. If any two-body is at thermal equilibrium there is no transfer of heat between them.
Case 1: The first two masses get mixed;
[tex]\rm mC_1 (T_1-T) + mC_2 (T_2-T) = 0\\\\ \rm C_1 (T_1 -T) + C_2 (T_2 -T) = 0\\\ \\ \rm C_1 (14-16) + C_2 (25-16) = 0\\\\ \rm -2 C_1 + 9 C_2 = 0\\\\C_1 = 4.5 C_2[/tex]
Case 2: The second and third masses get mixed:
[tex]\rm mC_2 (T_2 -T) + mC_3 (T_3 -T) = 0 \\\\ \rm C_2 (T_2 -T) + C_3(T_3 -T) = 0 \\\\ \rm C_2 (25 -27.9) + C_3 (33 -27.9) = 0\\\\ \rm -2.9 C_2 + 5.1 C_3 = 0 \\\\ \rm C_2 = 1.76 C_3[/tex]
By substituting the values we get
[tex]C_1= 4.5 (1.76 C_3)\\C_1 = 7.91 C_3[/tex]
Case 3: The first and third masses get mixed:
[tex]\rm mC_1 (T_1 -T) + m C_3 (T_3 -T) = 0\\\\ \rm C_1 (T_1 -T) + C_3 (T_3-T) = 0\\\\ \rm (7.91 C_3) (14 -T) + C_3 (33-T) = 0\\\\ \rm (7.91) (14-T) + 33-T = 0\\\\ \rm 111 -7.91T + 33 -T = 0\\\\ \rm 8.91T = 144\\\\ \rm T = 16.1^0 C[/tex]
Hence the equilibrium temperature when equal masses of the first and third are mixed will be 16.1° C.
To learn more about the thermal equilibrium refer to the link;
https://brainly.com/question/2637015
