Units and Dimensions 25 32.174 lb ft0.453 59 24 kg 0.3048 m (Ibf) Ibf s? X. Ibf 1 lb ft (0.0254)² m² in (in?) 2. in kg m = 6894.75 6894.75 N/m? - S m (Table 2.7 gives 6.894 757 × 10' as the value for...

Hello, can someone help me please explain why Celsius equals Kelvin in the solution shown in attached image?thank you

Extracted text: Units and Dimensions 25 32.174 lb ft0.453 59 24 kg 0.3048 m (Ibf) Ibf s? X. Ibf 1 lb ft (0.0254)² m² in (in?) 2. in kg m = 6894.75 6894.75 N/m? - S m (Table 2.7 gives 6.894 757 × 10' as the value for the conversion factor for lbf to pascal.) in? EXAMPLE 2.4 In the SI system, thermal conductivity has the unit W/(m K). The thermal xQ conductivity of a solid material can be calculated as k = where Q is the rate of heat ΑΔΤ transfer, x is the thickness of solid, A is the area of heat transfer and AT is the temperature difference across the solid. The following values were obtained experimentally: Q = 10 000 kJ/h, A = 1 m2, x = 100 mm and AT = 800 K. %3D %3D %3D (a) Calculate the thermal conductivity of the solid in W/(m K). (b) Express the thermal conductivity in kcal/(h m °C). (c) If thermal conductivity of a second material is 0.15 Btu/(h ft °F), which one will make a better thermal insulator? Solution 1000 J 10 000 kJ /(a) The rate of heat transfer, Q=10000 kJ/h =- kJ = 2777.78 I/s 3600 s h We have: x = 100 mm = 0.1 m, A = 1 m², AT = 800 K %3D Substituting these values in the given equation, we get xQ_0.1x 2777.78 k = ΑΔΤ 1x 800 J 0.347 sm K W = 0.347 m K %3D W 0.347-"= 0.347 m K J (b) sm K cal 4.1868 J cal = 0.347 3 298.557- °C hm°C S 3600 s x m × K × K