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+
+// Theory and Problems of Thermodynamics
+// Chapter 5
+// Second law of Thermodynamics
+// Example 7
+
+clear ;clc;
+
+//Given data
+T1 = 300                    // ambient atmosphere temperature in K
+T2 = 600                    // reservoir temperature in K
+TR = 240                    // refrigerator temperature in K
+
+
+// Calculations
+n = 1 - T2/T1
+// n = W/Q1 => W = 0.5*Q1
+// Q2 = Q1-W = 0.5*Q1
+
+COP_R = TR/(T1-TR)
+
+//COP_R = QL/W
+//QL = 4*W = 4*(0.5*Q1) = 2*Q1
+//QH = QL+W = 2*Q1 + 0.5*Q1 = 2.5Q1
+// Energy rejected to the ambient temperature 
+// Q2 + QH = 0.5*Q1+2.5*Q1 = 3*Q1
+// Hence, (Q2 + QH)/Q1 = 3
+
+mprintf('The ratio of energy rejected by both the devices to the ambient atmosphere \n to the energy absorbed by the engine from the reservoir at 600K is 3')
+
+
+
+