# Why does efficiency of heat engine not find by adding up two source value? Actually I have two questions.

1. we know the efficiency of heat engine equal {1-(t/T)},where t is sink temperature and T is source temperature. Now, if two source and one sink reservior are connected by a heat engine(as shown in picture),then if T is taken as sum of two source temperature and that gives wrong answer.

From picture, t=200 and T=800+400=1200 ,so efficiency is {1-(200/1200)} give wrong answer.

why do we need to find individually? But also adding up two individual efficiency give wrong answer(100% up efficiency).

If i take heat (from picture Qa+Qb) and work(net work divided by total heat supplied) instead of temperature then it gives me right answer.

1. why does the use of heat give me right answer?
• The version in terms of heat is the definition of efficiency. This just reduces to the equation in terms of temperature for a simple case like a Carnot engine. Other cases, not necessarily. Jul 25, 2020 at 18:07

Energy (including energy transport caused by a temperature difference, known as heat) is an extensive variable. In other words, if you duplicate a system and its input and output conditions, then all the energy terms will double. Linear superposition of energy fluxes of $$\dot{Q}_A$$ and $$\dot{Q}_B$$, if valid, will produce a flux of $$\dot{Q}_A+\dot{Q}_B$$.
In contrast, temperature is an intensive variable. If you duplicate a system at temperature $$T$$, the new temperature of the dual system is not $$2T$$ but generally remains $$T$$. Thus, the effective temperature of an 800 K reservoir and a 400 K reservoir used in conjunction is unlikely to be 1200 K.
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