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NEET - 2017

A carnot engine having an efficiency of110th of heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is -

(a) 1 J

(b) 90 J

(c) 99 J

(d0 100 J 

(b) Consider schematic diagram for a carnot engine as shown below. 

       

 In case of engine,

engine efficiency=work heat absorbed =Wq1

   Wq1=110

  10 Jq1=110

or 

when this engine is reversed, it takes in work W and heat q2 from cold reservoir and rejects 100 J of heat to hot reservoir.

 

     W+q2=q1  10+q2=100or    q2=90J

 

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NEET - 2016

 

The temperature inside a refrigerator is t2C and the room temperature is t1C . The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be -

(a)t1t1-t2      (b) t1+273t1-t2

(c) t2+273t1-t2    (d) t1+t2t1+273

 

(b) For a refrigerator, we know that 

             Q1W=Q1Q1-Q2=T1T1-T2

where,

Q1=amount of heat delivered to the room

W = electrical energy consumed

T1= room temperature= t1+273

T2=temperature of sink=t2+273

   Q11=t1+273t1+273-t2+273

Q1=t1+273t1-t2

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NEET - 2016

A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal = 4.2 Joules)

(a)23.65W
(b)236.5W
(c)2365W
(d)2.365W

(b) Given temperature of source T=30°C=30+273 T1=303K
Temperature of sink T2=4°C=4+273 T2=277K

As we know that 
Q1/Q2=T1/T2
=>Q2+W/Q2=T1/T2 ..........(W=Q1-Q2)

where Q2 is the amount of heat drawn from the sink (at T2),W is workdone on working substance,
Q1 is amount of heat rejected to source (at room temperature T1).

=>WT2+T2Q2=T1Q2
=>WT2=T1Q2-T2Q2
=>WT2=Q2(T1-T2)
=>W=Q2(T1/T2-1)
=>W=600X4.2X(303/277-1)
W=600X4.2X(26/277)
W=236.5Joules

Power=Workdone/Time=W/t=236.5/1=236.5W

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NEET - 2015

The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20°C, the temperature of the surroundings to which it rejects heat is

(a)31°C

(b)41°C

(c)11°C

(d)21°C

Key Concept:

Coefficient of performance (β) of a refrigerator is defined as the ratio of quantity of heat removed per cycle  to the work done on the working substance per cycle to remove this heat.

Given, coefficient of performance of a refrigerator β=5

Temperature of surface i.e. inside freezer,

T2=-20°C=-20+273=253K

Temperature of surrounding i.e. heat rejected outsider T1=?

So,β=T2/T1-T2

=>5=253T1-253

=>5T1=253/T1-253

=>5T1
-1265=253 =>

5T1=1518T1=15185=303.6K

T1=303.6-273=31°C

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A Carnot engine working between 300 K and 600K has work output of 800 J per cycle. What is amount of heat energy supplied to the engine from source per cycle ?

(1) 1800 J/cycle

(2) 1000 J/cycle

(3) 2000 J/cycle

(4) 1600 J/cycle

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If the door of a refrigerator is kept open, then which of the following is true ?

(1) Room is cooled

(2) Room is heated

(3) Room is either cooled or heated

(4) Room is neither cooled nor heated

(2) In a refrigerator, the heat dissipated in the atmosphere is more than that taken from the cooling chamber, therefore the room is heated if the door of a refrigerator is kept open.

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A Carnot's engine used first an ideal monoatomic gas then an ideal diatomic gas. If the source and sink temperature are 411°C and 69°C respectively and the engine extracts 1000 J of heat in each cycle, then area enclosed by the PV diagram is -

(1) 100 J

(2) 300 J

(3) 500 J

(4) 700 J

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The temperature of sink of Carnot engine is 27°C. Efficiency of engine is 25%. Then temperature of source is -

(1) 227°C

(2) 327°C

(3) 127°C

(4) 27°C

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The temperature of reservoir of Carnot's engine operating with an efficiency of 70% is 1000K. The temperature of its sink is -

(1) 300 K

(2) 400 K

(3) 500 K

(4) 700 K

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