👤

.c = 2090 J/Kg*K
m= 100g
t1 = -20 grade C
t_{t} = 0 grade C
Yt ( lamda ) = 33,5 * 10⁴ J/Kg
Q_{t} = ?


Răspuns :

Q=c·m·Delta t
delta t= t1 - t finala= 20°C-0°C =0°C
m=100g=0,1kg
Q=2090 J/kg°C · 0,1kg·20°C =
kg cu kg se simplifica,°C cu °C la fel se simplifica si ramane =4180 J