Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 13636 by chux last updated on 21/May/17

$$\mathrm{A}\mathrm{steam}\mathrm{envine}\mathrm{of}\mathrm{efficiency}70\mathrm{\%}$$$$\mathrm{burns}20\mathrm{g}\mathrm{of}\mathrm{coal}\mathrm{to}\mathrm{produce}10\mathrm{kJ}$$$$\mathrm{of}\mathrm{energy}.\mathrm{If}\mathrm{it}\mathrm{burns}200\mathrm{g}\mathrm{of}\mathrm{coal}$$$$\mathrm{per}\mathrm{second},\mathrm{calculate}\mathrm{its}\mathrm{output}$$$$\mathrm{power}.$$$$$$$$$$$$$$

Answered by ajfour last updated on 21/May/17

$$20g\to 10kJ$$$$200g\to 100kJ$$$$P=\frac{100kJ}{1s}=100kW.$$

Commented by chux last updated on 21/May/17

$$\mathrm{thanks}\mathrm{but}\mathrm{the}\mathrm{textbook}\mathrm{says}70\mathrm{kw}$$

Commented by sandy_suhendra last updated on 22/May/17

$$\mathrm{continued}$$$${\mathrm{P}}_{\mathrm{input}}=100\mathrm{kW}$$$${\mathrm{P}}_{\mathrm{output}}=\eta .{\mathrm{P}}_{\mathrm{input}}=0.7\times 100=70\mathrm{kW}$$

Commented by chux last updated on 22/May/17

$$\mathrm{thanks}\mathrm{boss}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com