224 Aptitude Test Problems in Physics
Hence we can determine T:
T — Tr2Tout1
— Trarout2
q, — 60 °C.
Tr2+ TOUtl- TOUt2- ri
2.17. The total amount of heat q liberated by the
space object per unit time is proportional to its
volume: q = aR 3 , where a is a coefficient. Since
the amount of heat given away per unit surface
area is proportional to T 4 , and in equilibrium, the
entire amount of the liberated heat is dissipated
into space, we can write q = 13R 2 T 1 (the area of
the surface is proportional to R^2 , and (3 is a coef-
ficient). Equating these two expressions for q,
we obtain
cc
T 4 = R.
13
Consequently, the fourth power of the temperature
of the object is proportional to its radius, and
hence a decrease in radius by half leads to a de-
crease in temperature only by a factor of V2
1.19.
2.18*. For definiteness, we shall assume that the li-
quid flowing in the inner tube 2 is cooled, i.e. T1 2 >
Tr2, and hence T 11 < T1 1. Since the cross-sec-
tional area 2S — S = S of the liquid flow in
the outer tube 1 is equal to the cross-sectional
area S of the liquid flow in the inner tube 2, and
their velocities coincide, the decrease in tempera-
ture of the liquid flowing in tube 2 from the entrance
to the exit is equal to the decrease in temperature
of the liquid flowing in tube 1. In other words,
the temperature difference in the liquids remains
constant along the heat exchanger, and hence
Tit Tft =-- Tf2 Tit• (1)
In view of the constancy of the temperature
difference, the rate of heat transfer is constant
along the heat exchanger. The amount of heat Q
transferred from the liquid flowing in tube 2 to
the liquid flowing in tube 1 during a time t isQ = Siatkt (T12 — T11).^ (2)