JEE MainPhysicsModerateThermodynamics

A coal based thermal power plant producing electricity operates
between the temperatures 27ºC and227ºC. The plant works at 80 of its
maximum theoretical efficiency. Complete burring of 1 kg of coalyields
3600 KJ of heat. A house needs 10 units of electricity each day. Coal
used for supplying theamount of energy for the house in one year is,d 1 kg iw.kZ tyus ij 3600 KJ mRiUu gksrh,d o"kZ rd nsus

Show answer and solution

Maximum efficiency of engine

= \left( 1 - \frac{T_{2}}{T_{1}} \right) \times 100 = 40\%

= \left( 1 - \frac{T_{2}}{T_{1}} \right) \times 100 = 40\%

\therefore

overall efficiency = 0.4 x 0.8 = 0.32

\therefore

dqy n{krk = 0.4 x 0.8 = 0.32
Requirement of house in one year,d lky esa ?kj
= 10 × 365 units = 3650 kwh = 3650 × 3600 kJ

\therefore

in put =

\frac{3650 \times 3600}{0.32}

\therefore

coalrequired =

\frac{\ In\ put\ }{36000}

=1141kg

Source: Grass-G / Ph-26 Physics paper 1 15 Jan.docx

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