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The efficiency of carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of the sink will be : -A. 100 KB. 600 KC. 400 KD. 500 K |
Answer» Correct Answer - C % n =`(1-T_2/T_1)xx100` For 50% `50/100 =1-500/T_1 rArr T_1=1000 K` For 60% `60/100`=1-`T_2/1000 rArr T_2`=400 K |
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