In a Carnot engine when `T_(2) = 0^(@)C` and `T_(1) = 200^(@)C` its efficiency is `eta_(1)` and when `T_(1) = 0^(@)C` and `T_(2) = -200^(@)C`. Its efficiency is `eta_(2)`, then what is `eta_(1)//eta_(2)`?

A Carnot's engine is made to work between 200^(@)C and 0^(@)C first and then between 0^(@)C and -200^(@)C . The ratio of efficiencies of the engine in the two cases is

A Carnot's engine is made to work between 200^(@)C and 0^(@)C first and then between 0^(@)C and -200^(@)C . The ratio of efficiencies of the engine in the two cases is

A carnot engine operating between temperature T_(1) andT_(2) has efficiency (1)/(6) . When T_(2) is lowered by 62 K, its efficiency increases to (1)/(3) . Then find the values of T_(1) and T_(2) .

A cannot engine with efficiency eta operates between two heat reservoirs with temperatures T_(1) and T_(2) , where T_(1) gt T_(2) . If only T_(1) is changed by 0.4% , the change in efficiency is Delta eta_(1) , whereas if only T_(2) is changed by 0.2% , the efficiency is changed by Delta eta_(2) . The ratio (Delta eta_(1))/(Delta eta_(2)) is approximately.

The efficiency of an engine is 1/2. When T_(2) is reduced by 100^(@)C , the efficiency becomes 2/3. Find T_(1)