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

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

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Hello, can someone help me please explain why Celsius equals Kelvin in the solution shown in attached image?thank you

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
Transcribed Image 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
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