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O sfera din platina are raza de 5 cm si temperatura de 95°C,se scufunda in 2 kg apa la temperatura de 4°C.
Care este temperatura de echilibru? Se da S=22000kg/m3
caldura specifica=120j/kg•gram
_
V=4| | r la a 3.a /supra 3

Urgent!!!!

Răspuns :

KostelMD

Se da:

Platină

p=22000kg/m3

c1=120J/kg**C

r=5cm=0,05m

t1=95*C

apa

m2=2kg

t2=4*C

V=4*pi*r^3/3

c2=4185J/kg**C

___________

tf-?

m1=pV=p*4*pi*r^3/3=22000*4*3,14*125*10^-6/3=11,51kg

Qcedat=Qprimit

c1m1(t1-tf)=c2m2(tf-t2)

c1m1t1-c1m1tf=c2m2tf-c2m2t2

c1m1t1+c2m2t2=c2m2tf+c1m1tf

tf(c2m2+c1m1)=c1m1t1+c2m2t2

tf=(c1m1t1+c2m2t2)/(c2m2+c1m1)=164694/9751,2=16,88*C

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