Animated Solution for Physics - Thermodynamics: A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

Imagine you are an engineer tasked with upgrading a power plant. The plant operates on the principles of a Carnot engine, the theoretical limit of thermodynamic perfection. Your goal? To boost the efficiency of the engine from a modest $40\%$ to a much more impressive $60\%$. But there is a catch: the cooling river (the sink) that absorbs the exhaust heat cannot be changed. How hot must you make the furnace (the source) to achieve this? Let's embark on this thermodynamic journey.

A Carnot engine is an idealized heat engine that operates between two thermal reservoirs: a hot source at temperature $T_{\text{source}}$ and a cold sink at temperature $T_{\text{sink}}$. The engine absorbs heat from the source, converts a portion of it into useful work, and dumps the remaining heat into the sink.

The efficiency $\eta$ of this engine tells us what fraction of the input heat is successfully converted into work. The master key to solving any Carnot engine problem is its efficiency formula:

Crucial Rule: In thermodynamics, temperatures must always be plugged into formulas in Kelvin (absolute temperature). Using Celsius will lead to disastrously wrong ratios!

Decoding the Sink Temperature

In our first scenario, the engine is operating at an efficiency of $40\%$. Converting this to a decimal, we get $\eta =0.4$. We are also told that the source temperature is $500\text{ K}$.

Let's substitute these known values into our master equation to uncover the hidden temperature of the sink:

Now, we perform a simple algebraic rearrangement. We move the temperature fraction to the left side and the $0.4$ to the right side:

To isolate $T_{\text{sink}}$, we multiply $0.6$ by $500$:

So, the exhaust is being dumped into a sink maintained at $300\text{ K}$ (which is roughly room temperature, $27^{\circ }\text{C}$).

Boosting the Efficiency

Now comes the upgrade phase. The boss wants the efficiency bumped up to $60\%$, meaning our new ${\eta }^{\prime }=0.6$. The constraint given in the problem is that the exhaust (sink) temperature remains exactly the same. Therefore, our sink is still sitting at $300\text{ K}$.

We need to find the new intake (source) temperature, let's call it $T_{\text{source}}^{\prime }$. We set up the equation for this new scenario:

Once again, we rearrange the terms to isolate the fraction containing our unknown variable:

To find $T_{\text{source}}^{\prime }$, we swap it with the $0.4$:

Calculating this final division gives us:

To achieve a $60\%$ efficiency without changing the cooling system, you must crank up the furnace temperature to a blazing $750\text{ K}$. The physics here is beautiful: to extract more work from the same cold environment, you must inject heat from a much hotter source, thereby increasing the thermal gradient that drives the engine.