Question

the efficiency of an engine is 70 percent .if the temperature of sink is decreased by 10 poercent its efficiency becom 80 percent . find t2 and t1

Answer 1

Pliz write full questions

Answer 2

The question is incorrect. If the temperature of the sink is decreased by 10 percent, the resulting efficiency will become 73%, and the temperatures won't be deducible.

The correct data seems to be: The temperature of the sink is decreased by 10K. This case is considered.

Answer: $T_{1} = 100K, T_{2} = 30K$

Explanation:

Let $T_{1}$ and $T_{2}$ be the temperatures of source and sink respectively. Then, efficiency is given by:

$\eta = \frac{T_{1} - T_{2}}{T_{1}} = 1 - \frac{T_{2}}{T_{1}}$

So, initially:

$1 - \frac{T_{2}}{T_{1}} = \frac{7}{10}$

Or, $\frac{T_{2}}{T_{1}} = \frac{3}{10}$

Now, if $T_{2}$ becomes $T_{2} - 10$, the efficiency shall be:

$\eta = 1 - \frac{(T_{2} - 10)}{T_{1}} = \frac{8}{10}$

Or, $\frac{(T_{2} - 10)}{T_{1}} = \frac{2}{10}$

Solving these equations, you'll get:

$T_{2} = 30K, T_{1} = 100K$

Note: if the convention is taken opposite, that is, 1 is the sink and 2 the source, then the answer would invert too. By invert, I do not mean 1/answer.