Answer:
a). V = 14.6208 litres : b). = 0.29844 Litres/sec
Step-by-step explanation:
a).
Step 1.
Volume when one half of rod is inserted = (50/2)*(2.2) = 55cm³
Step 2.
As we know that,
V = (ρV
IΔT)
/ (ρL
)
where
= 290 - 4.2 = 285.8K
sepecififc heat of aluminum = 0.215 cal/g
density = 2.7 g/cm3
= (2.7)*(55)*(0.215)*(285.8) / (0.125)*(20900)*(1/4186)
V = 14.6208 litres
b).
Step 1.
As we know that
P = kA(dT/dx) = 31*2.2*[(290-4.2)/25]
= 779.6624 W
Step 2.
Now, approximate boil rate of helium is:
P/ρL= [(779.6624)*(10^3g/kg)] / [(0.125)*(20900)]
= 0.29844 Litres/sec