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Assume that under certain conditions the heat loss of an object is proportional to the exposed surface area. Relate the heat loss of a cubic object with side length 6 in. to one with a side length of 12 in. Now, consider two irregularly shaped objects, such as two submarines. 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. Relate the energy needed in the actual submarine to that required by the scaled model. Specify the assumptions you have made.

Answer :

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

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