The cat is fat. The cat’s fur and the cat’s fat provide the same amount of insulation. How much kitty blubber is equivalent to 1.0 cm of fur with regards to heat transfer by conduction?Please put it in this formatting. Thermal conductivities of some commonsubstancesSubstanceSilverCopperAluminumWoodConcreteFiberglassStyrofoamFurBlubberk (W/m. K)4063852050.12 -0.040.80.040.0270.040.24 A. Draw a picture or diagram of the problem. Label relevant vectors, objects, and coordinates. (½ pt)Significant Figures : (½ pt)LB. List the relevant known quantities for the problem. (½ pt)| C. Make a guess as to the approximate result you wouldexpect for the unknown variable. (pt)D. Mathematical calculations starting from the basic relevantformulas and ending with a boxed answer. (7 pts)E. Brief written explanations of each step in box D. (½pt)

Given tha The thickness of the fur is ∆x1=1cm We know that The thermal resistance is given by Rθ =∆xAxk Where Rθ is the absolute thermal resistance in k/W across the thickness of the sample ∆x is the thickness of the sample k is thermal conductivity of the sample A is the cross sectional area Now from the above formula we can write Rθ α∆xk Here they asked to find the cat’s fur and the cat’s fat provide the same amount of insulation Then ∆x1k1=∆x2k2 Now thermal resistance is given by ∆x2 =∆x1xk2k1 From the above table we can take values of blubber (0.24W/m.k) and fur (0.04W/m.k) ∆x2 =∆x1xk2k1=1cmx0.240.04= 6cm Therefore from the above calculations we can say that 6cm blubber is equivalent to 1cm of cat fur