Question

intr un calorimetru se gasesc: m1=2,5 kg de apă la t1=40°C si 500 g apa la t2=5°C. Care este temperatura de echilibru? Se cunoaste ca=4180 j/kg*k

Answer 1

m₂=500 g=0,5 kg

Qced=m₁c(t₁-t)

Qprim=m₂c(t-t₂)

Qced=Qprim<=>m₁c(t₁-θ) =m₂c(θ-t₂)<=>2,5*4180(40-θ)=0,5*4180(θ-5)|:4180=>2,5*40-2,5θ=0,5θ-0,5*5<=>100+2,5=(0,5+2,5)θ<=>102,5=3θ=>θ=102,5/3=34,16 °C

Answer 2

Qcedat=Qprimit

cedeaza apa1

primeste apa2

m1=2.5 kg

t1=40 C

m2=0.5 kg

t2=5 C

ca=4180 j\kgk

Θ=?

Qcedat= ca x m1 x (40-Θ)

Qprimit= ca x m2 x (Θ-5)

ca x m1 x (40-Θ)=ca x m2 x (Θ-5)

4180j\kgk x 2.5 kg x (40-Θ)=4180j\kgk x 0.5kg x (Θ-5)

10450j\k x (40-Θ) = 8360j\k x (Θ-5)

10450j\k x 40 - 10450j\k x Θ = 8360j\k x Θ - 8360j\k x 5

418000j\k-10450 x Θ= 8360j\k x Θ - 41800j\k

418000j\k+41800j\k=8360j\k x Θ + 10450j\k x Θ

459800j\k=Θ(8360j\k+10450j\k)

459800j\k=18810j\k x Θ

Θ=459800j\k : 18810j\k

Θ=24.44j\k

Succes!