Answer:
factor of 100 more heat loss in actual
Step-by-step explanation:
Given:
- Length of the cube = 6 in
- Length of another cube = 12 in
- Actual submarine size = 70 ft
- Scale model = 7-ft
Find:
Relate the heat loss of a 70-ft submarine to that of a 7-ft scale model. Suppose you are interested in the amount of energy needed to maintain a con- stant internal temperature in the submarine.
Solution:
- The surface area of the small cube is 6 · 36.
- Scaling the sides by k means the new surface area is 6 · (6k · 6k) = k
^2*
(6 · 36).
- Therefore, double the lengths increases the surface area 4 times (and so there will also be 4 times as much heat loss).
- Now consider two irregularly shaped objects, such as submarines. If we
can measure heat loss from a 7 foot scale model, heat is lost from a 70 foot sub is:
- An increase by a factor of 10 will mean that heat loss (which is proportional to surface area) increases by a factor of 10^2 = 100
