Question

A Styrofoam cooler has outside dimensions of B = 84.0 cm, W = 47.0 cm, and H = 59.0 cm. The thickness of each wall of the cooler, t, is 8.0 cm. Styrofoam has a density, rho, of 1.0 kg/m3. (a) What is the volume of the Styrofoam used in cubic inches? (b) What is the mass in lbm? (c) How many gallons of liquid could be stored in the cooler?

(a) The volume of the Styrofoam used is 14214.37 in3.
(b) The mass is 9.6 lbm.
(c) The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

Answer 1

The volume of the Styrofoam used is 14214.37 in³. The mass is 9.6 lbm. The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

Step-by-step explanation:

To find the volume of the Styrofoam used in cubic inches, we need to convert the outside dimensions from centimeters to inches, subtract the volume of the cooler, and divide by the thickness of each wall. The volume of the Styrofoam used is 14214.37 in³.

To find the mass in lbm, we need to convert the density from kg/m³ to lbm/in³ and multiply by the volume of the Styrofoam used. The mass is 9.6 lbm.

To find the number of gallons of liquid that could be stored in the cooler, we need to convert the volume of the cooler from cm³ to gallons. The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

Answer 2

a)
`V_s=4989.7895\ in^3`

b)
`mass=0.1803\ lb`

c)
`V_i=39.93\ gallons`

Step-by-step explanation:

• external breadth of cooler,
  `B=84\ cm`

• external width of cooler,
  `W=47\ cm`

• external height of cooler,
  `H=59\ cm`

• ∵ thickness of each wall is,
  `t=8\ cm`

Therefore,

• internal breadth of cooler,
  `B_i=76\ cm`

• internal width of cooler,
  `W_i=39\ cm`

• internal height of cooler,
  `H_i=51\ cm`

a)

External volume of the structure:

`V=B.H.W`

`V=84*47*59/ 2.54^3`

`V=14214.3828\ in^3`

Internal volume of the structure:

`V_i=B_i.H_i.W_i`

`V_i=76* 39* 51/ 2.54^3`

`V_i=9224.5932\ in^3`

∴Volume of Styrofoam used:

`V_s=V-V_i`

`V_s=14214.3828-9224.5932`

`V_s=4989.7895\ in^3`

b)

given that density of Styrofoam,
`\rho=1\ kg.m^(-3)=3.613* 10^(-5)\ lb.in^(-3)`

we know,

`\rm mass= density * volume`

`mass=4989.7895* 3.613* 10^(-5)`

`mass=0.1803\ lb`

c)

Volume of liquid it can hold
`=V_i`

`V_i=39.93\ gallons`