# A heat engine with an efficiency of 17 % does 836 J of work in each cycle. How much heat must be supplied from the high-temperature source in each cycle?

Mar 16, 2018

$4917.65 J$

#### Explanation:

Coefficient of efficiency of a heat engine is defined as the ratio of work done $W$ by it to heat added from heat source$Q$.

So,we can write, $\frac{17}{100} = \frac{W}{Q}$

Given, $W = 836 J$

So, $Q = 836 \cdot \left(\frac{100}{17}\right) = 4917.65 J$

Mar 16, 2018

See below
$4920 J$

#### Explanation:

Let's recall the definition of efficiency:
$E = {W}_{o u t} / {W}_{\text{in}} \cdot 100$

So, we are given the output of work here as $836 J$ and the efficiency as 17%

Let's substitute and solve:
${W}_{o u t} / E \cdot 100 = {W}_{\text{in}}$
$\frac{836 J}{17} \cdot 100 \approx 4920 J$

Mar 16, 2018

The total energy consumed per cycle is 4918 J

#### Explanation:

17 % efficiency means that in one cycle, $\text{work done"/"total energy consumed} = 0.17$

Therefore if 836 J of work is done in each cycle, solving the following equation for $\text{total energy consumed}$ will provide the answer to this question.

$\frac{836 J}{\text{total energy consumed}} = 0.17$

$\text{total energy consumed} = \frac{836 J}{0.17} = 4918 J$

I hope this helps,
Steve