238 Aptitude Test Problems in Physics
be the length of the lower part of the duct where
the cast iron is molten, and /^2 be the length of
the upper part where the cast iron is in the sol-
id phase. Then we can write the condition of
the constancy of heat flux in the form (the cross-
sectional area of the duct is constant)XIII] (Ti- Tmeit) xsol (Tmelt— T2)
ll^ is^where xiici and x 501 are the thermal conductivities
of the liquid and solid cast iron. Considering that
xllq = kx501, we obtain/I— /2k (Ti— Tmelt)
Tmelt — T2
The total length of the duct is 1 1 + / 2. Therefore,
the part of the duct occupied by the molten metal
is determined from the relation/ 1 k(Ti--Tmelt)
1 1+ 12 k(Ti— Tmelt)+ (Tmelt — T2)
2.35*. The total amount of heat Q emitted in
space per unit time remains unchanged since it is
determined by the energy liberated during the
operation of the appliances of the station. Since
only the outer surface of the screen emits into space
(this radiation depends only on its temperature),
the temperature of the screen must be equal to
the initial temperature T = 500 K of the station.
However, the screen emits the same amount of heat
Q inwards. This radiation reaches the envelope
of the station and is absorbed by it. Therefore,
the total amount of heat supplied to the station
per unit time is the sum of the heat Q liberated
during the operation of the appliances and the
amount of heat Q absorbed by the inner surface
of the screen, i.e. is equal to 2Q. According to
the heat balance condition, the same amount of
heat must be emitted, and henceQ (^) T4
2Q =Tx