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 .