Question

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width 8.75 inches, length 8.75 inches, height 23 inches.
The hot coffee has a density of 5.8 oz/in3 and the cold coffee has a density of 3.5 oz/ in3

(A) What is the volume of the coffee in each tank?
Show all math work needed to complete this problem.

(B) What is the mass of the coffee in each the tank?
Show all math work needed to complete this problem.


(C) How many tanks can be placed on a table which can support a maximum weight of 800 lb?
Explain why or why not.
Show all math work needed to complete this problem.

Answer 1

Answer:

Part A

Dimensions of tank:

• width = 8.75 in
• length = 8.75 in
• height = 23 in

⇒ Volume of each tank = width × length × height

                                       = 8.75 × 8.75 × 23

                                       = 1760.9375 in³

Part B

Given:

• Hot coffee density = 5.8 oz/in³
• Cold coffee density = 3.5 oz/in³

mass = density × volume

⇒ mass of hot coffee = 5.8 oz/in³ × 1760.9375 in³

                                     = 10213.44 oz (nearest hundredth)

⇒ mass of cold coffee = 3.5 oz/in³ × 1760.9375 in³

                                      = 6163.28 oz (nearest hundredth)

Part C

Convert the masses of the tanks from ounces to pounds

1 lb = 16 oz

⇒ mass of hot coffee tank = 10213.44 ÷ 16

                                            = 638.34 lb (nearest hundredth)

⇒ mass of cold coffee tank = 6163.28 ÷ 16

                                             = 385.21 lb (nearest hundredth)

If the table can hold a maximum of 800 lb:

hot coffee:  800 lb ÷ 638.34 lb = 1.25 (nearest hundredth)

cold coffee:  800 lb ÷ 385.21  lb = 2.08 (nearest hundredth)

Therefore, either:

• 1 tank of hot coffee   OR
• 2 tanks of cold coffee

can be placed on a table which can support a maximum weight of 800 lb