(9%) Problem 11: Consider a car radiator at 113.5°C in a 41.5°C environment and the exchange of heat between them by radiation.Calculate the net rate of heat transfer, in watts, by radiation from the car's radiator, if it has an emissivity of 0.750 and a 1.4-m? surface area.Grade SummaryDeductionsPotentialQ/At =0%100%7895 6sin()cos()tan()SubmissionsHOMEAttempts remaining: 10(0% per attempt)detailed viewcotan()asin()acos()E A 4atan()acotan()sinh()123tanh)cosh()ODegreescotanh()ENDRadiansVol BACKSPACECLEARDELSubmitHintFeedbackI give up!Hints: 0% deduction per hint. Hints remaining: 1Feedback: 0% deduction per feedback.

The net rate of heat transferQΔt.given,emmisivity(e)=0.750surface area(A)=1.4 m2Initial temperature(T1)=113.5 oC=113.5+273.15=386.65 Kfinal temperature(T2)=41.5 oC=41.5+273.15=314.65 KStefan-Boltzmann constant(σ)=5.67×10-8 Wm2K4The rate of heat transferQΔt=eAσT24-T14substituting values,rate of heat transferQΔt=0.7501.4 m25.67×10-8 Wm2K4314.65 K4-386.65 K4or, rate of heat transferQΔt=-747.03 Wso,The net rate of heat transferQΔt=-747.03 W.

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Problem II: Consider a car radiator at 113.5°C in a 41.5°C environment and the exchange of heat between them by radiation. Calculate the net rate of heat transfer, in watts, by radiation from the car's radiator, if it has an emissivity of 0.750 and a 1.4-m² surface area. Grade Summar Deductions Potential PAt =

Problem II: Consider a car radiator at 113.5°C in a 41.5°C environment and the exchange of heat between them by radiation. Calculate the net rate of heat transfer, in watts, by radiation from the car's radiator, if it has an emissivity of 0.750 and a 1.4-m² surface area. Grade Summar Deductions Potential PAt =

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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