Question

An emergency response unit is preparing for a hurricane (and the potential for a prolonged power outage). As part of their preparation, they are filling Yeti Tundra 110 coolers with ice. The inside dimensions of each cooler is approximately 15 inches high, eleven inches wide and 30 inches long. Based on these dimensions, the volume capacity of each cooler is 4,900 cubic inches and the external surface area of the four sides plus top and bottom is given as 1,800 square inches. The sides, top and bottom of the Yeti coolers is made of 2-inch polyurethane with an R-value of 14 (ft²-°F-hr/BTU).

If each cooler is being filled with 84 pounds of ice and the average outside temperature is expected to be 81°F, how many days will it take for all of the ice to melt? (For your calculations, assume that the temperature inside the coolers is 32°F and that the coolers remain closed.)

Answer 1

Ans) Let us first estimate the heat loss(Q) through the cooler, by using relation,

Q = U x A x T

where, U = Thermal conductivity = 1/R value

A = Surface area of cooler = 1800 sq.in ir 12.5 sq.ft

  T = Tempearture difference

Given, R-value of polyurethane = 14 F hr/BTU

=> U-value = 1 / 14 BTU / F hr

Putting values,

=> Q = (1/14) x 12.5 x (81 - 32)

=> Q = 43.75 BTU/hr

Amount of ice in cooler = 84 lb

We know, heat of fusion of water = 144 BTU/lb which means 144 BTU of heat is required to melt 1 lb of ice

Hence, heat required to melt 84 lb ice = 144 BTU/lb x 84 lb = 12096 BTU

Amount of time required to melt all ice = Total heat / heat loss = 12096 BTU/43.75 BTU/hr = 276.48 hrs

=> Number of days required to melt all of the ice = 276.48 hrs / 24 hrs/day = 11.52 days

Hence, about 11.52 days required to melt all of the ice