Question

intrun cazan se introduc 12 kg de apa cu temp de 80 de grade celsius si 27 de kg de apa la 15 grade. Aflati temperatura finala .
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Answer 1

Răspuns: temperatura de echilibru termic tₓ = 35°C

Explicație:

pornim de la ideea ca apa calda se raceste si apa rece se incalzeste pana la o temperatura de echilibru tₓ  si nu sunt pierderi de caldura

Q cedat = Q primit

Q cedata de apa calda = m₁c ( t₂ - tₓ )

Q primita de apa rece  = m₂ c (tₓ - t₁)

m₁c ( t₂ - tₓ )  =  m₂ c (tₓ - t₁)    se simplifica c

12  ( 80 - tₓ) = 27 ( tₓ - 15 )

960 - 12 tₓ = 27 tₓ - 405

960  + 405 = 39 tₓ

tₓ = 1365 : 39 = 35

Answer 2

$m_1=12\ kg \\ T_1=80^oC \\ m_2=27\ kg \\ T_2=15^oC \\ c_1=c_2=c \\ Q=mc\Delta T\\Q_1=Q_2\\\Rightarrow m_1c(T_1-T_f)=m_2c(T_f-T_2)\\\Rightarrow12\ kg \cdot (80^oC-T_f)=27\ kg\cdot(T_f-15^oC)\\\Rightarrow4(80-T_f)=9(T_f-15)\\\Rightarrow320-4T_f=9T_f-135\\\Rightarrow-4T_f-9T_f=-135-320\\\Rightarrow -13T_f=-455 \\ \Rightarrow T_f=\frac{455}{13}^oC \\\Rightarrow \boxed{T_f=35^oC}$