Thermodynamics

If minimum possible work is done by a refrigerator in converting 100 gm of water at 0°C to ice ….

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Q: If minimum possible work is done by a refrigerator in converting 100 gm of water at 0°C to ice , how much heat (in calorie) is released to the surroundings at temperature 27 °C (Latent heat of ice = 80 cal/gram) to the nearest integer ?

Click to See Solution :
Ans: (8791)

Sol: T1 = 27 °C = 300 K , Lice = 80 cal/g

Heat energy extracted from water , Q2 = m L

Q2 = 100 x 80 = 8000 cal

T2 = 0 °C = 273 K

$\displaystyle \frac{Q_1}{Q_2} = \frac{T_1}{T_2}$

$\displaystyle Q_1 = Q_2 \times \frac{T_1}{T_2} $

$\displaystyle Q_1 = 8000 \times \frac{300}{273} $

= 8791 cal

 

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work ) …

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Q: Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work ) between temperatures T1 and T2 . The temperature of the hot reservoir of the first engine is T1 and temperature of cold reservoir of the second engine is T2 . T is the temperature of sink of first engine which is also the source for the second engine . How is T related to T1 and T2 , if both the engines perform equal amount to work ?

(a) $\displaystyle T = \sqrt{T_1 T_2}$

(b) T = 0

(c) $\displaystyle T = \frac{2T_1 T_2}{T_1 + T_2}$

$\displaystyle T = \frac{T_1+ T_2}{2}$

Click to See Solution :
Ans: (d)
Sol: For first engine ,

$\displaystyle W_1 = Q_1 – Q_2 $

and $\displaystyle \frac{Q_2}{Q_1} = \frac{T}{T_1} $

$\displaystyle W_1 = Q_1[1 – \frac{Q_2}{Q_1} ] $

$\displaystyle W_1 = Q_1[1 – \frac{T}{T_1} ] $ ..(i)

For Second engine ,

$\displaystyle W’ = Q_2 – Q_2′ $

and $\displaystyle \frac{Q_2′}{Q_2} = \frac{T_2}{T} $

$\displaystyle W_2 = Q_2 [1-\frac{Q_2′}{Q_2}]$

$\displaystyle W_2 = Q_2 [1-\frac{T_2}{T}]$ …(ii)

Work done by first engine(W_1) = work done by second engine (W_2)

$\displaystyle Q_1[1 – \frac{T}{T_1} ] = Q_2 [1-\frac{T_2}{T}] $

$\displaystyle \frac{Q_1}{Q_2} = \frac{1-\frac{T_2}{T}}{1-\frac{T}{T_1}}$

$\displaystyle \frac{T_1}{T} = \frac{1-\frac{T_2}{T}}{1-\frac{T}{T_1}} $

$\displaystyle T_1 – T = T – T_2 $

$\displaystyle T = \frac{T_1 + T_2}{2}$

 

Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C, when the cooling water is available at -35.7 °C. then, choose the correct option(s).

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Q: Consider a heat engine working with a hot reservoir of the furnace gases at 2100 °C , when the cooling water is available at -35.7 °C. then, choose the correct option(s).

(a) Engine efficiency can be made more than 90% without altering operating temperatures

(b) Engine efficiency always remains less than 90 %

(c) Engine efficiency is much lower practically due to irreversibility in the actual cycle

(d) Engine efficiency is 90% but such temperatures are not available practically

Click to See Answer :
Ans: (c)

Sol: Temperature of hot Reservoir , T1 = 2100 + 273 = 2373 k

Temperature of cold Reservoir , T2 = -35.7 + 273 = 237.3 K

Maximum Efficiency , $\displaystyle \eta_{max} = 1 – \frac{T_2}{T_1}$

$\displaystyle \eta_{max} = 1 – \frac{237.3}{2373}$

$\displaystyle \eta_{max} = 1 – \frac{1}{10} = \frac{9}{10}$

Percentage efficiency $\displaystyle = \frac{9}{10} \times 100$

= 90 %

 

During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Then, choose the correct option(s).

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Q: During a process a system receives 30 kJ of heat from a reservoir and does 60 kJ of work. Then, choose the correct option(s).

(a) It is possible to reach initial state by a quasistatic isothermal process

(b) It is possible to reach initial state using a constant volume process

(c) It is possible to reach initial state using a constant pressure process

(d) It is possible to reach initial state using an adiabatic process

Click to See Answer :
Ans: (d)
Sol: Here ΔQ = 30 kJ , ΔW = 60 kJ

For Process 1-2 ,

ΔQ1-2 = ΔU1-2 + ΔW1-2

30 = ΔU1-2 + 60

ΔU1-2 = – 30 k J

For Process 2-1 ,

ΔQ2-1 = ΔU2-1 + ΔW2-1

0 = U1 – U2 + ΔW2-1 (for Adiabatic Process , ΔQ = 0 )

0 = 30 + ΔW2-1

ΔW2-1 = – 30 kJ

Therefore , by doing a work of 30 kJ on the System , it can go back to Initial state

 

A heat engine receives heat at rate of 1500 kJ/min and gives an output of 8.2 kW. Rate of heat rejection by the engine is

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Q: A heat engine receives heat at rate of 1500 kJ/min and gives an output of 8.2 kW. Rate of heat rejection by the engine is

(a) 1500 kJmin-1

(b) 8.2 kJs-1

(c) 16.8 kJs-1

(d) 8.2 kJmin-1

Click to See Answer :
Ans: (c)
Sol: Heat received per sec = 1500/60 = 25 kJ/sec

Work output per sec = 8.2 kW = 8.2 kJ/sec

Rate of heat rejection by engine = 25 – 8.2 = 16.8 kJ/sec