43. Let the mass of the steam be m
s
and that of the ice be m
i
. Then L
F
m
c
+ c
w
m
c
(T
f
− 0.0
◦
C) =
L
s
m
s
+ c
w
m
s
(100
◦
C
− T
f
), where T
f
= 50
◦
C is the final temperature. We solve for m
s
:
m
s
=
L
F
m
c
+ c
w
m
c
(T
f
− 0.0
◦
C)
L
s
+ c
w
(100
◦
C
− T
f
)
=
(79.7 cal/g)(150 g) + (1 cal/g
·
◦
C)(150 g)(50
◦
C
− 0.0
◦
C)
539 cal/g + (1 cal/g
·C
◦
)(100
◦
C
− 50
◦
C)
=
33 g .