42. The power consumed by the system is

P

=

1

20%

cm∆T

t

=

1

20%

(4.18 J/g

·

◦

C)(200

× 10

3

cm

3

)(1 g/cm

3

)(40

◦

C

− 20

◦

C)

(1.0 h)(3600 s/h)

=

2.3

× 10

4

W .

The area needed is then

A =

2.3

× 10

4

W

700 W/m

2

= 3 3 m

2

.