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Read the passage carefully and answer the followingquestions. Imagine a system, that can keep the room temperature within a narrow range between 20^(@)C to 25^(@)C. the system includes a heat engine operating with variable power P =KT, where K is a constant coefficient, depending upon the thermal insulation of the room, the area of the walls and the thickness of the walls. T is temperature of the room in degree, when the room temperature drops lower than 20^(@)C, the engine turns on,when the temperature increase over 25^(@)C, theengineturnsoff, room looses energy at a ate of K(T-T_(0) is the outdoor temperature. Thde heat capacity of the room is C. Given (T_(0)=10^(@)C,In((3)/(2))=0.4,In((6)/(5))=0.18,(C)/(K)=750SI-unit) Suppose at t = 0, the engine turns off, after how much time interval, again, the engine will turn on: |
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Answer» 10 minute `(dT)/(dt) = -(K)?(C)(T-T_(0))` `rArr`` int_(25)^(20)(dT)/(T-10)=-(k)/( C ) int_(0)^(1) dt` `rArr`` In(10)/(15) = -(K)?(C)t` `rArr`` In (3)- In(2) = (K)/( C )t` `rArr`` t= 0.4 xx ( C ) /(k) = 5 MIN` When engine turns on `(dT)/(dt) =-( K )/( C ) (T-T_(0))+(P)/( C )` `rArr``- ( K)/( C ) [ -T+T_(0)+3T]` `rarr`` int_(20)^(25) (dT)/(21T +T_(0)) = (+Kt)/( C )` `rArr`` In((60)/(50)) = (K)/( C )t` `rArr``t=( C ) /(2K) In((6)/(5))` =1.125 min |
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