Do wody o masie 10kg i temperaturze 20°C wlano wodę o masie 20kg i temperaturze 80°C. Jaka będzie temperatura końcowa wody po wymieszaniu?
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Dane:
m₁ = 20 kg
T₁ = 80⁰C
m₂ = 10 kg
T₂ = 20⁰C
T = ?
Rozw.:
Q₁ = m × Cw × ΔT
ΔT= T₁ -T
Q₁ = m₁ × Cw × ( T₁ -T)
Q₂ = m₂ × Cw × ΔT
ΔT= T -T₂
Q₂ = m₂ × Cw × ( T -T₂ )
Q₁ = Q₂
m₁ × Cw × ( T₁ -T) = m₂ × Cw × ( T -T₂ )
m₁ × ( T₁ -T) = m₂ × ( T -T₂ )
m₁T₁ - m₁T = m₂T - m₂T₂
m₁T₁ + m₂T₂ = m₂T + m₁T
m₁T₁ + m₂T₂ = T(m₂ + m₁)
T = m₁T₁ + m₂T₂ / m₂ + m₁ = 20 kg × 80⁰C + 10kg × 20⁰C/10 kg + 20kg = 60⁰C