A fiberglass box with dimensions of 24 inch long 18 inch wid

A fiberglass box with dimensions of 24 inch long. 18 inch wide and 14 inch in height, with a thickness of 1 inch contains ice at a temperature of 492 (Ra) Rankine. The temperature outside is 580.6 (Ra) Rankine. If the coefficient of fiberglass is K=0.04 W/m-L. What is the Heat Flow Rate (Q/t) into the box? How much ice is melted in 15 hours?

Solution

SOlution

Using the equaiton

                 dQ / dt = K A dT / dx

                 Here dQ / dt = heat flow rate

                     dT = temperature different

                      Area of cuboid box =A = 2wL + 2Lh + 2wh = 2(24 * 18) + 2(24* 14) + 2(18 * 14) = 2040inch²

                       Thickness = dx = 1inch

                     dQ / dt = 0.04 W / m .K (2040)inch² ( 580.6- 492) / inch

                        dQ/dt = 0.04 W / m .K (2040)inch² (88) K / inch

                                   = 0.04 W / m K ( 1.31m²) (88K) / 0.0254 m

                                    = 181.54 W

         2) In 15h the amount of ice melt would be

                          heat of fusion of water is

                       Q = m Hf

              Here Hf is latent heat of fusion

                       Amount of ice would melt will be found by using the relation

                             dQ / dt = m Hf / dt = 181.54W

                                             m = (181.54 J / s ) (15 * 3600s) / 334 J / g

                                             m = 29351g

                                             m=29.35kg

This would be the amount of ice melt.