∆T=80-30
∆T=50 ARE ACEEASI
VALOARE INDIFERENT DE GRADE FIE CELSIUS FIE KELVIN
n=V/Vm
n=400/22,4
n=17,857 moli
1mol.........................213kcal
17,857.......................Q
Q=213X17,857
Q=3803,541Kcal
Se consuma doar
60x3803,541/100=2282,124Kcal
Q=mc∆T
2282,124x1000cal=mx4185x50
Cum nu au aceeasi
unitate de masura se transforma cal in Jouli
2282,124x1000cal...............................y
1cal.......................................................4,186J
Y=9539,28X1000J
Abia acum se pot
face calcule matematice
9539,28X1000=mx4185x50
m=9539,28X1000/4185x50
m=45,58 grame apa
