Question

During a high fever, a 60.0-kg-patient's normal metabolism increases by 10.2%. This results in an increase of 10.2% in the heat given off by the person. When the person slowly walks up five flights of stairs (20.0 m), she normally releases 95.0 kJ of heat. What is the ratio of the patient's efficiency efever when she walks up the stairs with a fever to her efficiency ehealthy when she walks up the stairs healthy without a fever. e fever ehealthy

Answer 1

The answer for above problem is explained below.

Given that Q = 95 kJ And Eo = mgh = 60 kg x 9.81 m/s2x zom Eo = 11.772 KJ z Energy De Total required by the person is, E = FotQ = 11772 J + 95000 J = 106772J efficiency of a person is, healthy n = = Eo a 11972] 11025 10 6772 J And Qcevee 95000 J x 110.2 2104690: 100 Total energy, E = Eo t Q Fever E = 11772 J + 104690J E= 116462 J Y= @seves - Bon - 1,2732, = 0.10108] 11772 T 10,10108 116462J esever = 0.10108 = 10.916825 | e healthy Ool1025