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One of the most efficient engines ever developed operates between 2100 K and 700 K. Its actual efficiency is 40%. What percentage of its maximum possible efficiency is this? |
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Answer» Solution :`T_(1) = 2100K, T_(2)= 700 K` The maximum possible efficiency `= eta = (T_(1) - T_(2))/( T_(1)) ` `= (2100 - 700)/(2100) = 0.667` `= 66.7%` ACTUAL efficiency `= 40%` `("Actual efficiency")/(" Maximum efficiency") = ( 40 xx 100)/(100 xx 66. 67) = 60%` |
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