Question

A coal based thermal power plant producing electricity operates between the temperatures $ {27^\circ }C $ and $ {227^\circ }C $ . The plant works at $ 80\% $ of its maximum theoretical efficiency. Complete burning of $ 1kg $ of coal yields $ 3600KJ $ of heat. A house needs $ 10\,units $ of electricity each day. Coal used for supplying the amount of energy for the house in one year is:
A. $ 1141kg $
B. $ 580kg $
C. $ 605kg $
D. $ 765kg $

Answer 1

Hint: To solve the given question, first we will convert the given units in joule as per the comfortability to solve the further steps. And then we will find thermal efficiency. After that we will assume the amount of coal as $ n $ and solve for it.

Complete step by step solution:
Electrical energy requirement of a house: $ 10\,units $ $ = 10kWhr = 10 \times 3.6 \times {10^6}J $
Electrical Energy requirement of the house per year:
 $ 365 \times 3.6 \times {10^7} = 1.314 \times {10^7}kJ $
Thermal efficiency of the heat engine: $ 0.4 \times 0.8 = 0.32 $
Let the amount of coal required be $ n $ , then
 $
  n \times 0.32 \times 36000 = 1.314 \times {10^7} \\
   \Rightarrow n = \dfrac{{1.314 \times {{10}^7}}}{{11520}} = 1140.625 \\
  \therefore n \approx 1141kg \\
 $
Therefore, Coal used for supplying the amount of energy for the house in one year is $ 1141kg $ .
Hence, the correct option is A. $ 1141kg $ .

Note:
The thermal efficiency, $ n $ , represents the fraction of heat that is converted to work. It is a dimensionless performance measure of a heat engine that uses thermal energy, such as a steam turbine, an internal combustion engine, or a refrigerator.