232 Aptitude Test Problems in Physics
amount of heat AQ lost by the vessel leads to a
decrease in the temperature of the vessel by A Ty„.
For the vessel with water, we obtain
(Mwcw mvescves) A Tves ,
where Mw and cw are the mass and specific heat of
water, mves and eves are the relevant quantities
for ihe vessel.
For the vessel with water and the ball, we obtain
AQ (^2) = (Mwcw mwcw mbc b) Alves,^
where mb and cb are the mass and specific heat of
the ball. By hypothesis, mves << Mw and mb =
M. Besides, eves << cw, and hence we can write
AQ 1 = Mwcw A T„,es, AQ^2 = Mw (cw cb) A Tves.
It can easily be seen that the change A Tves of tem-
perature in the two vessels occurs during different
time intervals At, and At 2 so that
Arves
F (ryes— n Mwcw
ATves CC
F (Tves—T) Mw (cw + cb)
Hence we obtain
Ott cw
At2 cw cb
Therefore, the following relation will be fulfilled
for the total times t 1 and t 2 of cooling of the vessels:
t2 cw --^1 cb -^ k
ti cw^
,
whence cb/c.,,, = k — 1.
2.29. If the level of water in a calorimeter has
become higher, it means that a part of water has
been frozen (the volume of water increases during
freezing). On the other hand, we can state that some
amount of water has not been frozen since otherwise
its volume would have increased by a factor of
A t2