Răspunsuri

2014-04-21T09:16:42+03:00
Q=mcΔt
Δt=t-to
m=0,2kg
Q=66960J
Δt=Q/m*c
c apa=4185J/kg*K
Δt=66960J/0,2KG*4185J/Kg*K
Δt=66960J/837J/kg*K
Δt=80K
Δt=t-to
t=100C= 373,15 K
to=373,15-80=293,15K=20 grade C



ce e to?
temperatura initiala a apei
multumesc
cu placere
2014-04-21T11:07:42+03:00
Temperatura de fierbere a apei este 100 C
Deci apa s-a incalzit cu Δt=100-t
Q=mcΔt=mc(100-t)
Q=100mc-mct
t=(Q-100mc)/mc
Q=6696j
m=0,2kg
c=4182j/kgK (tabel)
inlocuiesti si calculezi!
Succes!