Solutions (^233)
Pw/Pice^ 1.1, and the level of water in the calo-
rimeter would have increased by (h/3) (1.1 — 1)
2.5 cm, while by hypothesis Ah = 0.5 cm. Thus,
the temperature established in the calorimeter
is 0 °C.
Using this condition, we can write
cw mw, (Tw, — 0 °C)
= —1 Am + cicemice (0 °C — Tice),
where Am is the mass of frozen water, and Tice
is the initial temperature of ice. As was men-
tioned above, the volume of water increases as
a result of freezing by a factor of pw/p ice, and hence
MS.= (—ear— —1) Am (2)
Pice
where S is the cross-sectional area of the calorim-
eter. Substituting Am from Eq. (2) into Eq. (1)
and using the relations mw (h/3) NS and mice =
(h/3) Pic eS, we obtain
cw S pw T w
= —XS
Ah PicePw h
cicePiceSTIce T.
Pw — Pice
Hence
X 3Ah (^) Pw
Tice=
cite Pw —Pice
cw pw
T wti — 54 °C.
ice Pice
2.30*. (1) Assuming that water and ice are incom-
pressible, we can find the decrease in the tempera-
ture of the mixture as a result of the increase in the
external pressure:
AT= ( 21 / - ) X 1 K ny. 0.18 K.
Such a small change in temperature indicates that
only a small mass of ice will melt, i.e. Am < mice•
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