Question

Do 0,5kg. wody o temeraturze 20`C dolano 1/3 litra wody i temperaturze 60`C.Oblicz temperature końcową wody.

Proszę pomóżcie szybko!!!!!!!!!!

Answer 1

m1=0,5kg

m2=1/3kg

T1=20 C

 

T2=60 C

Tk=temp. koncowa

 

m1*T1=m2*T2 bilans cieplny

m1*(Tk-T1)=m2*(T2-Tk)

m1Tk-T1m1=m2T2-Tkm2

m1Tk+Tkm2=m2T2+T1m1

Tk(m1+m2)=m2T2+T1m1

Tk=m2T2+T1m1/(m1+m2)

 

Tk=1/3*60 + 0,5*20 / 1/3+0,5

Tk=20+10/5/6

Tk=30/5/6

Tk=36 C

 

Mam nadzieje ze pomoglam pozdrawiam :)

 

p.s. napewno dobrze bo wlasnie tez to mam na fizyce ;) 

Answer 2

dane:

m1 = 0,5 kg

t1 = 20*C

m2 = 1/3 kg  (1 l wody = 1 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 = (0,5kg*20C + 1/3kg*60C)/(0,5kg+1/3kg) = 30kg*C/(5/6 kg)

tk = 36 *C

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