Problem 1 [4 marks] You are working for a camping gear manufacturer and your bos

Question

Problem 1 [4 marks] You are working for a camping gear manufacturer and your boss has asked you to design a new portable cooler. The portable cooler will have the shape of a rectangular prism, with a height of 25 cm, a width of 20 cm and a length of 35 cm. The walls of the cooler will be composed of a thin inner liner (negligible resistance to heat transfer), a gap containing water, and an outer insulating wall having a thickness of 20 mm and a thermal conductivity of 0.025 Wm K1. The idea is to place the cooler inside a freezer before a long trip so that the water trapped inside the wall freezes. The contents could then be maintained at a low temperature because the melting ice would be able to absorb a relatively large quantity of heat transferred from the warm surroundings. Assuming that you would like the ice to take approximately 5 hours to melt on a typical day, how thick would you make the ice layer? You may neglect all multi-dimensional heat transfer effects, and all radiation effects. You may also assume that on a typical day the outside surface temperature of the cooler would be maintained at 30°CThe latent heat of fusion of ice can be assumed to be 334.0 kJ/kg and the density of the ice is 916.7 kg/m

Explanation / Answer

Applying conductivity equation across any surface of the cooler.

Q/t = kA(T2-T1)/d and assuming ice temperature 0 DegC

consider a surface of height 25 and width 20 cm; area = 25*20 = 500 cm2 or 0.05 m2.

Q = 0.025*0.05*(30-0)/0.02 = 1.875 W or 1.875 J/s

Now on a typical day, heat will have o be asobe by ice at above mentioned rate,

heat absorbed in 5 hour = 5*3600*1.875 = 33750 J = 33.75 kJ

mass required of ice for absorbing above heat = 33.75/334 = 0.101 kg of ice

now by using ice density, volume of ice required = 0.101/916.7 = 1.102x10-4 m3 = 110.2 cm3.

Now since we are doing our calulations for a surface of height 25 and width 20 cm corresponding to area of 500 cm2, the thickness of ice would be = 110.2/500 = 0.22 cm = 2.2 mm

Please note that we can repeat the all of above calculations by considering a different surface as well but the final result would come same.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.