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NEET - 2017

A carnot engine having an efficiency of$\frac{1}{10}$th 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=$\frac{work}{heatabsorbed}=\frac{W}{{q}_{1}}$

$\therefore \frac{W}{{q}_{1}}=\frac{1}{10}$

$\Rightarrow \frac{10J}{{q}_{1}}=\frac{1}{10}$

or

when this engine is reversed, it takes in work W and heat ${q}_{2}$ from cold reservoir and rejects 100 J of heat to hot reservoir.

$\therefore W+{q}_{2}={q}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow 10+{q}_{2}=100\phantom{\rule{0ex}{0ex}}or{q}_{2}=90J$

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NEET - 2016

The temperature inside a refrigerator is ${{t}_{2}}^{\circ}C$and the room temperature is ${{t}_{1}}^{\circ}C$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be -

(a)$\frac{{t}_{1}}{{t}_{1}-{t}_{2}}$ (b) $\frac{{t}_{1}+273}{{t}_{1}-{t}_{2}}$

(c) $\frac{{t}_{2}+273}{{t}_{1}-{t}_{2}}$ (d) $\frac{{t}_{1}+{t}_{2}}{{t}_{1}+273}$

(b) For a refrigerator, we know that

$\frac{{Q}_{1}}{W}=\frac{{Q}_{1}}{{Q}_{1}-{Q}_{2}}=\frac{{T}_{1}}{{T}_{1}-{T}_{2}}$

where,

${Q}_{1}$=amount of heat delivered to the room

*W* = electrical energy consumed

${T}_{1}$= room temperature= ${t}_{1}+273$

${T}_{2}=$temperature of sink=${t}_{2}+273$

$\therefore $ $\frac{{Q}_{1}}{1}=\frac{{t}_{1}+273}{{t}_{1}+273-\left({t}_{2}+273\right)}$

$\Rightarrow {Q}_{1}=\frac{{t}_{1}+273}{{t}_{1}-{t}_{2}}$

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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 T_{2}=4°C=4+273 T_{2}=277K

As we know that

Q_{1}/Q_{2}=T_{1}/T_{2}

=>Q_{2}+W/Q_{2}=T_{1}/T_{2} ..........(W=Q_{1}-Q_{2})

where Q_{2} is the amount of heat drawn from the sink (at T_{2}),W is workdone on working substance,

Q_{1} is amount of heat rejected to source (at room temperature ${T}_{1}$).

=>WT_{2}+T_{2}Q_{2}=T_{1}Q_{2}

=>WT_{2}=T_{1}Q_{2}-T_{2}Q_{2}

=>WT_{2}=Q_{2}(T_{1}-T_{2})

=>W=Q_{2}(T_{1}/T_{2}-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,

T_{2}=-20°C=-20+273=253K

Temperature of surrounding i.e. heat rejected outsider T_{1}=?

So,β=T_{2}/T_{1}-T_{2}=>$5=\frac{253}{{T}_{1}-253}$

=>5T_{1=253/T1-253=>$5{T}_{1}$}-1265=253 =>_{$5{T}_{1}$}_{=$1518{T}_{1}$=$\frac{1518}{5}$=$303.6K$${T}_{1}=303.6-273=31\xb0C$}

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A Carnot engine working between 300 *K* and 600*K* 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*

(4) $\eta =\frac{{T}_{1}-{T}_{2}}{{T}_{1}}=\frac{W}{Q}$ ⇒ $Q=\left(\frac{{\displaystyle {T}_{1}}}{{\displaystyle {T}_{1}-{T}_{2}}}\right)\text{\hspace{0.17em}}W$

$=\frac{600}{(600-300)}\times 800$ =1600 *J*

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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 *

(3) $\eta =1-\frac{{T}_{2}}{{T}_{1}}=1-\frac{(273+69)}{(273+411)}=0.5$

⇒ Work done $=\eta \times Q=0.5\times 1000=500\text{\hspace{0.17em}}J$

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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*

(3) $\eta =1-\frac{{T}_{2}}{{T}_{1}}$ ⇒ $\frac{25}{100}=1-\frac{300}{{T}_{1}}$ ⇒ $\frac{1}{4}=1-\frac{300}{{T}_{1}}$

${T}_{1}=400\text{\hspace{0.17em}}K=127\xb0C$

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

(1) 300 *K*

(2) 400 *K *

(3) 500 *K*

(4) 700 *K *

(1) $\eta =1-\frac{{T}_{2}}{{T}_{1}}$ ⇒ $\frac{70}{100}=1-\frac{{T}_{2}}{1000}$

⇒ *T*_{2} = 300 *K*

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