When 4200J of heat are added to a 0,15m-long silver bar, its length increases by 4,3*10^-3m. What is the mass of the bar?
SVOLGIMENTO:
[math]DT= Dl/(l_0 \cdot K)= [(4,3 \cdot 10^-3)/(0,15 \cdot 19 \cdot 10^-6)]°C= 1508,8°C[/math]
[math]m=Q/(c \cdot DT)= [4200/ (235 \cdot 1508,8)]Kg= 0,012Kg[/math]
