If we consider solar system consisting of the earth and sun only as of the ideal thermodynamic system. The sun works as source of energy having temperature 6000 K and earth as sink having temperature 300 K, the efficiency of solar system would be on the basis of exchange of radiations

Question

If we consider solar system consisting of the earth and sun only as of the ideal thermodynamic system. The sun works as source of energy having temperature 6000 K and earth as sink having temperature 300 K, the efficiency of solar system would be on the basis of exchange of radiations (A) 30% (B) 65% (C) 75% (D) 95%

Tardigrade Admin · Accepted Answer

Temperature of source T1=6000 K Temperature of sink T2=300 K Then efficiency η =1-(T2/T1)=1-(300/6000)=(19/20) Efficiency percentage =(19/20)× 100=95%

Q. If we consider solar system consisting of the earth and sun only as of the ideal thermodynamic system. The sun works as source of energy having temperature 6000 K and earth as sink having temperature 300 K, the efficiency of solar system would be on the basis of exchange of radiations

30%

65%

75%

95%

Temperature of source $T_{1} = 6000 K$ Temperature of sink $T_{2} = 300 K$ Then efficiency $η = 1 - \frac{T _{2}}{T _{1}} = 1 - \frac{300}{6000} = \frac{19}{20}$ Efficiency percentage $= \frac{19}{20} \times 100 = 95$

Q. If we consider solar system consisting of the earth and sun only as of the ideal thermodynamic system. The sun works as source of energy having temperature 6000 K and earth as sink having temperature 300 K, the efficiency of solar system would be on the basis of exchange of radiations

30%

65%

75%

95%

Temperature of source T1​=6000K Temperature of sink T2​=300K Then efficiency η=1−T1​T2​​=1−6000300​=2019​ Efficiency percentage =2019​×100=95

Temperature of source $T_{1} = 6000 K$ Temperature of sink $T_{2} = 300 K$ Then efficiency $η = 1 - \frac{T _{2}}{T _{1}} = 1 - \frac{300}{6000} = \frac{19}{20}$ Efficiency percentage $= \frac{19}{20} \times 100 = 95$