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// Exa 20.2
clc;
clear all;
// Given data
M=200;// mass in grams
Heat =1;//Sp. Heat of water(cal/gm degree)
T1=30;//Initial temperature (degree Celsius)
T2=40;//Final temperature (degree Celsius)
// Solution
Pwr_rad=4.18*M*Heat*(T2-T1); // in Watt
printf(' The power radiated = %.2f kW \n',Pwr_rad/1000);
// The answer in the textbook is mentioned as 8.3 kW but by calculation it comes out to be 8.36 kW.
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