Thermodynamics

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Height of stairs h= 10m

Energy produced by burning 1 kg of fat = 7000Kcal

Energy produced by burning 5kg of fat = 5 $\times 7000=35000Kcal$

Energy utilised in going up and down one time

= mgh + $\frac{1}{2}mgh$ = $\frac{3}{2}mgh$

= $\frac{3}{2}\times 60\times 10\times 10$

= 9000J= 9000/4.2=3000/1.4cal

Number of times, the person

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Temperature of the source T₁= 500K and sink T₂= 300K

Work done W= 1000J

Efficiency of Carnot engine = 1-T₂/T₁= 1-300/500= 200/500= 2/5

Efficiency = W/Q₁

So Q₁= W/efficiency = 1000 $\left(\frac{5}{2}\right)=2500J\mathit{}$

This is a short answer type question as classified in NCERT Exemplar

During driving temperature of the gas increases while volume remains constant. So according charle's law, at constant volume V.

Pressure is directly proportional to temperature. Therefore pressure of gas increases.

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Yes during adiabatic compression the temperature of a gas increases while no heat is

In adiabatic compression dQ=0

From the first law of thermodynamics dU= dQ-dW

dU=-dW

in compression work is done on the gas i.e work done is negative

dU=

This is a short answer type question as classified in NCERT Exemplar

If a refrigerator's doors is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.

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For path1

Heat Q₁= 1000J

Work done =W₁

For path 2

Work done W₂= W₁-100

As change in internal energy is same

dU=Q₁-W₁=Q₂-W₂

1000-W₁=Q₂-W₁+100

Q₂= 1000-100= 900J

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Yes this is possible when the entire heat supplied to the system is utilised in expansion.

So its working against the surroundings.

This is a long answer type question as classified in NCERT Exemplar

(a) Initially the piston is in equilibrium P_i=P_a

(b) On supplying heat , the gas expands from V_o to V_i

so increase in volume of the gas =V_i-V_o

as the piston is of unit cross sectional area hence extension in the spring

x= $\frac{V_i-V_0}{area}=V_i-V_0$

force exerted

This is a long answer type question as classified in NCERT Exemplar

Slope of the curve = f(V) , where V is the volume

Slope of P = f(V) curve at ((P_o, V₀ )= f(V_o)

Slope of adiabatic at (P_o, V₀ )= k(-Y)V_o^-1-Y =-YP_o/V_o

Now heat absorbed in the process P= f(V)

dQ=dU+dW= nC_vdT+pdV

pV=nRT

T= pV/nR

T= $\frac{1}{nR}\left[f\left(V\right)+V{f}^{\text{'}}\left(V\right)dV\right]$

$\frac{dQ}{dV}=$ nC_{v $\frac{dT}{dV}+p\frac{dV}{dV}=n{C}_V\frac{dT}{dV}+P$}

$\frac{C_V}{R}\left[f\right(V_o+V_0{f}^{\text{'}}\left(V\right)+f\left(V\right)]$

A

This is a long answer type question as classified in NCERT Exemplar

(a) For process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

 = nC_vdT= nC_v(T_B-T_A)

 = $\frac{3}{2}R\left(T_B-T_A\right)$

= $\frac{3}{2}\left(R{T}_B-R{T}_A\right)=\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

Heat exchanged = $\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

(b) For process BC , p =constant

dQ= dU+dW  = $\frac{3}{2}R\left(T_C-T_B\right)+P_B(V_C-V_B)$

heat exchanged = $\frac{5}{2}{P}_B\left(V_C{-V}_A\right)$

(c) For process CD , because C