The efficiency of a Carnot engine is eta when | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Rearranging this,we get $\frac{T_2}{T_1} = 1 - \eta$,or $\frac{T_1}{T_2

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$\frac{0.5 T_2 \eta}{(1-1.5 \eta)(1-\eta)}$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Rearranging this,we get $\frac{T_2}{T_1} = 1 - \eta$,or $\frac{T_1}{T_2} = \frac{1}{1 - \eta}$.To increase the efficiency to $1.5 \eta$ while keeping $T_2$ constant,let the new hot reservoir temperature be $T_1' = T_1 + \Delta T$.The new efficiency is $1.5 \eta = 1 - \frac{T_2}{T_1 + \Delta T}$.Rearranging,$\frac{T_2}{T_1 + \Delta T} = 1 - 1.5 \eta$.Taking the reciprocal,$\frac{T_1 + \Delta T}{T_2} = \frac{1}{1 - 1.5 \eta}$.Splitting the left side,$\frac{T_1}{T_2} + \frac{\Delta T}{T_2} = \frac{1}{1 - 1.5 \eta}$.Substituting $\frac{T_1}{T_2} = \frac{1}{1 - \eta}$,we get $\frac{1}{1 - \eta} + \frac{\Delta T}{T_2} = \frac{1}{1 - 1.5 \eta}$.Thus,$\frac{\Delta T}{T_2} = \frac{1}{1 - 1.5 \eta} - \frac{1}{1 - \eta} = \frac{(1 - \eta) - (1 - 1.5 \eta)}{(1 - 1.5 \eta)(1 - \eta)} = \frac{0.5 \eta}{(1 - 1.5 \eta)(1 - \eta)}$.Therefore,$\Delta T = \frac{0.5 T_2 \eta}{(1 - 1.5 \eta)(1 - \eta)}$.

The efficiency of a Carnot engine is $\eta$ when its hot and cold reservoirs are maintained at temperatures $T_1$ and $T_2$,respectively. To increase the efficiency to $1.5 \eta$,the increase in temperature $(\Delta T)$ of the hot reservoir,while keeping the cold reservoir constant at $T_2$,is

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