Do wanny zawierającej 2 kg wody o temp 20 stopni C dolano 4 kg wody o temp. 60 stopni C. Oblicz temperaturę końcową mieszaniny? Proszę o pomoc dość szybką :))
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dane:
m1 = 2 kg
t1 = 20*C
m2 = 4 kg
t2 = 60*C
szukane:
tk = ?
Q pobrane = Q oddane
m1C(tk-t1) = m2C(t2-tk) /:C
m1(tk-t1) = m2(t2-tk)
m1tk-m1t1 = m2t2-m2tk
m1tk+m2tk = m1t1+m2t2
tk(m1+m2) = m1t1+m2t2 /:(m1+m2)
tk = (m1t1+m2t2)/(m1+m2)
tk = (2kg*20*C+4kg*60*C)/(2kg+4kg) = 280/6 *C = 46,666 *C
tk ≈ 46,7*C
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