HSSC-I (11th Class) Physics Unit-10 Thermodynamics: Numerical Problems List (New Book/KPK Text Board, Peshawar) With Verified Answers

Numerical Problems 1: Water at 20°C falls from a height of 854 meters. If the whole energy is used in increasing the temperature, find out the final temperature. Specific heat of water is 4200 J⁻¹ K⁻¹ kg (22°C) 

... 

Given:

Initial temperature of water = T₁ = 20°C = 20°C + 273 = 293 K

Height of water = h = 854 m

Specific heat of water = C = 4200 J K⁻¹ kg⁻¹

Value of g = 9.81 m s⁻²

To Find:

Final temperature = T₂ = ?

Solution:

Using specific heat Equation

△Q = m C △T

and 

Whereas the heat gained △Q = gain in Potential Energy = m g h 

and  △T = T₂ - T₁  So

m g h  = m c (T₂ - T₁)

by separating the value of  T₂ we get 

T₂ = `\frac {gh}{c}` + T₁

by putting the corresponding values

T₂ = `\frac {9.81 m s⁻² ｘ 854 m}{4200 J K⁻¹ kg⁻¹}` + 293 K

T₂ = 1.99 K + 293 K

T₂ = 294.99 K

or

T₂ = 294.99 K - 273

T₂ = 21.99 °C

or

T₂ = 22°C ------------------------Ans 

Numerical Problems 2: 25,200 J of heat is supplied to the system while the system does 6,000 J of work. Calculate the change in internal energy of the system. (19200 J)

Ans. (1.63 msᐨ²)

... 

Given:

Heat supplied to the system = △Q = 25,200 J

Work done by the system = △W = 6,000 J

To Find:

Change in the internal energy of the system = △U = ?

Solution:

Using the First law of thermodynamics we have 

△Q = △U  + △W

or 

△U  = △Q - △W

by putting values 

△U  = 25,200 J - 6,000 J

△U  = 19,200 J  ----------------Ans.

Numerical Problems 3: A sample of ideal gas is uniformly heated at constant pressure. If the amount of 180 J of heat is supplied to the gas, calculate the change in internal energy of the gas and Work done by the gas. Take γ = 1.41 (127.66 J, 52.34 J)

... 

Given:

Heat supplied to the system at constant pressure  = △`\Q_p` = 180 J

γ = 1.41

To Find:

Change in the internal energy of the system = △U = ?

Work done by the system = △W = ?

Solution:

Since  at constant pressure we have an equation for molar specific heat.

△`\Q_p` = n `\C_p`△T  ------------(1)

and we also know that at constant volume  △`\Q_v` = △U

so 

△`\Q_v` = △ U = n `\C_v`△T -----------(2)

and the ratio of the Specific heat at constant pressure to Specific heat at volume is given by 

γ = `\frac {C_p}{C_v}` 

or

`\C_p` = γ `\C_v` ----------------(3)

Putting the value `\C_p` = γ `\{C_v}` in equation (1) 

△`\Q_p` = n γ `\C_v`△T

or

△`\Q_p` = γ n`\C_v`△T

As △ U = n `\C_v`△T 

Thus 

△`\Q_p` = γ △ U

or 

△ U  = `\frac {△Q_p}{γ}`

putting values we get

 △ U  = `\frac {180 J}{1.41}`

△ U  = 127.66 J -------------Ans. 1

Now using the First law of thermodynamics we have 

△Q = △U  + △W

or 

△W  = △Q - △U

by putting values 

△W  = 180 J - 127.66 J

△W  = 52.34 J  ----------------Ans. 2

Numerical Problems 4: Find the efficiency of a Carnot's heat engine working between the steam and ice points? (26.8%)

... 

Given:

Temperature of Steam ( HTR ) = T₁ = 100°C = (100°C + 273) K = 273 K

Temperature of ice ( LTR ) = T₂ = 0°C = (20°C + 273) K = 273 K

To Find:

Efficiency of heat engine = η = ?

Solution:

Efficiency of the heat Engine in term of temperature is  

η = (1 - `\frac {T₂}{T₁}`) 

by putting the corresponding values

η = (1 - `\frac {273 K}{293 K}`) 

η = (1 - 0.732) 

η = 0.268  

In Percentage

η = 0.268 ｘ  `\frac {100}{100}`

η = `\frac {26.8}{100}`

η = 26.8 % ----------------Ans. 

Numerical Problems 5: A Carnot heat engine absorbs 2000 J of heat from the source of heat engine at 227 °C and rejects 1200 J of heat during each cycle to sink. Calculate efficiency of engine temperature of sink and amount of work done during each cycle. (40%, 27°C, 800 J)

... 

Given:

Heat absorbed by the Carnot engine from heat source = Q₁ = 2000 J 

Temperature of HTR = T₁ = 227°C = (100°C + 273) K = 500 K

Heat rjected by the Carnot engine to sink (LTR) = Q₂ = 1200 J 

To Find:

Efficiency of Carnot engine = η = ?

Temperature of sink =  T₂ = ?

Work done in a cycle = △W = ?

Solution:

Efficiency of the heat Engine in term of heat is  

η = (1 - `\frac {Q₂}{Q₁}`)

by putting the corresponding values

η = (1 - `\frac {1200 J}{2000 J}`)

η = (1 - 0.6)

η = 0.4  

In Percentage

η = 0.4 ｘ `\frac {100}{100}`

η = `\frac {40.0}{100}`

η = 40 % ----------------Ans. 

Now to find the temperature of the sink (LTR), we have the formula for efficiency of the heat Engine in term of temperature is  

η = (1 - `\frac {T₂}{T₁}`) 

by separating T₂ we get

T₂ = (1 - η)T₁

by putting values

T₂ = (1 - 40%)500 K

T₂ = (1 - `\frac {40}{100}`)500 K

T₂ = (0.6)500 K

T₂ = 300 K

or

T₂ = 27°C --------------Ans. 2

Now work done 

△W = Q₁ - Q₂

△W = 2000 J - 1200 J

△W = 800 J -------------Ans. 3

Numerical Problems 6: What is the least amount of work that must be performed to freeze one gram of water at 0°C by means of a refrigerator? Take the temperature of the surrounding as 37°C. How much heat is passed on to the surrounding during this process? (45.54 J, 381.54 J)

... 

Given:

Mass of water = m = 1 g = 0.001 kg

Temperature of atmosphere (HTR) = T₁ = 37°C = (37°C + 273) K = 310 K 

Temperature of heat sink (LTR) = T₂ = 0°C = (0°C + 273) K = 273 K

To Find:

Work done  = △W = ?

Heat rejected to the surrounding = Q₁ = ?

Solution:

To find the work done by the refrigerator  to freeze a water we have

W = `\frac {Q₂}{E_{cool}}` -----------(1)

But Q₂ and `\E_{cool}` are also unkown. So, we have to first find the heat of sink (LTR) Q₂ and Coefficient of performance= `\E_{cool}`

To find heat of sink (LTR) Q₂ we have formula

Q₂ = m `\H_f`  

Latent heat of fusion of water `\H_f`  = 336,000 J/K

Q₂ = 0.001 kg 𝐱 336,000 J/Kg

Q₂ = 336 J

and to find Coefficient of performance= `\E_{cool}`we have

`\E_{cool}` = `\frac {310 K}{310 K - 273 K}`

`\E_{cool}` = 7.38

Now using the equation (1)

W = `\frac {336J}{7.38}`

W = 45.54 J ----------------Ans.1

Now to find Heat rejected to the surrounding by formula of refrigerator 

Q₁ = Q₂  + W

Q₁ = 336 J  + 45.54 J

Q₁ = 381.54 J--------------------Ans.2

Numerical Problems 7: Calculate the change in entropy when 10 kg of water is heated from 90°C to 100°C? (Specific heat of water is 4180 J mole⁻¹K⁻¹. (1135.8 JK) 

... 

Given:

Mass of water = m = 10 kg

Initial temperature of water = T₁ = 90°C = (90°C + 273) K = 363 K 

Final temperature of water = T₂ = 100°C = (100°C + 273) K = 373 K

Specific heat of water = C = 4180 J mol⁻¹ K⁻¹

To Find:

Change in Entropy  = △S = ?

Solution:

The formula for change in entropy is 

△S = `\frac {△Q}{T}`

So, we have to first find the △Q by using formula

△Q = m C △T

where   △T = T₂ - T₁ so

△Q = m C ( T₂ - T₁ )

△Q = 10 kg ｘ 4180 j mol⁻¹ K⁻¹ ( 373 K - 363 K)

△Q = 41800 kg J mol⁻¹ K⁻¹ ( 10 K)

△Q = 418000 J 

and

T = `\T_{ave}` = `\frac { T₁ + T₂}{2}`

T = `\frac {363 K3 + 373 K }{2}`

T = 368 K

Now

△S = `\frac {418000 J }{368 K}`

△S = 1135.87 J/K ----------------Ans

Numerical Problems 8: A system absorbs 1176 J of heat and at the same time does 352.8 J of external work. Find the change in internal energy of the system? Find the change in internal energy in the system when it absorbs 1050 J of heat while 84 J of work is done? What will be the change in internal energy of the gas if 210J of heat is removed at constant volume? (823.2 J, 966 J, -210 J)

... 

The question has three parts:

(1) 

A system absorbs 1176 J of heat and at the same time does 352.8 J of external work. Find the change in internal energy of the system?

Given:

Heat absorbed by the system = △Q = 1176 J

Work done by the system = △W = 352.8 J 

To Find:

Change in internal energy  = △U = ?

Solution:

Using the First law of thermodynamics we have 

△Q = △U  + △W

or 

△U  = △Q - △W

by putting values 

△U  = 1176 J - 352.8 J

△U  = 823.2 J  ----------------Ans.1

*********************************************

(2) 

Find the change in internal energy in the system when it absorbs 1050 J of heat while 84 J of work is done?

Given:

Heat absorbed by the system = △Q = 1050 J

Work done by the system = △W = 84 J 

To Find:

Change in internal energy  = △U = ?

Solution:

Using the First law of thermodynamics we have 

△U  = △Q - △W

by putting values 

△U  = 1050 J - 84 J

△U  = 966 J  ----------------Ans. 2

*******************************************

(3) 

What will be the change in internal energy of the gas if 210 J of heat is removed at constant volume?

Given:

Heat removed by the system = △Q = -210 J

Work done by the system = △W =0 ( due to at constant volume)

To Find:

Change in internal energy  = △U = ?

Solution:

At constant volume the first law of thermodynamics we have (△W = 0 J) so,

△U  = △Q 

by putting values 

△U  = 1050 J - 84 J

△U  = -210 J  ----------------Ans.3

Numerical Problems 9: An ideal gas at 20.0°C and a pressure of 1.50 x 10⁵ Pa is in a container having a volume of 1.00 L. (a) Determine the number of moles of gas in the container. (b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature. (6.16 x 10² mol, 395 K)

... 

Given:

Initial temperature of an ideal gas = T₁ = 20°C = (20°C + 273) K = 293 K 

Initial Pressure = P₁  = 1.50 ｘ 10⁵ Pa

Initial Volume = V₁  = 1 l  = 0.001 m³  as { 1 m³ = 1000 litters]

Final Volume = V₂ = 2 V₁ = 0.002 m³  

Value of R (universal gas constant) =R = 8.315 J mol⁻¹ K⁻¹

Pressure fall to P₂ = 1 atm = 1013.25 mb = 1013.25 hPa =  1013.25 ｘ 10²  Pa

To Find:

(a) Number of mole of gas = n = ?

(b) Final temperature  = T₂ = ? (given: when pressure fall P₂ = 1 atm = 1013.25 Pa)

Solution:

(a) using ideal gas equation

P₁ V₁  = n R T₁ 

or

n = `\frac {P₁ V₁}{R T₁}`

by putting values

n = `\frac {1.50 ｘ 10⁵ Pa ｘ  0.001 m³ }{8.315 J mol⁻¹ K⁻¹ ｘ 293 K }`

n = `\frac {1.50 ｘ 10² m³ Pa }{2436.295 J mol⁻¹ }`

n = 0.06156  mol

or

n = 6.156 ｘ 10⁻²  mol

***********************************

(b) using ideal gas equation

P₂ V₂  = n R T₂

or

T₂ = `\frac {P₂ V₂}{n R}`

by putting values

T₂ = `\frac {1013.25 ｘ 10² Pa ｘ  0.002 m³ }{6.156 ｘ 10⁻²  mol ｘ 8.315 J mol⁻¹}`

T₂ = `\frac {2.02 ｘ 10² m³ Pa }{0.5119 J K⁻¹ }`

T₂ = 394.6  K

or

T₂ = 395  K -----------------Ans.

Numerical Problems 10: A block of ice at 273 K is put in thermal contact with a container of steam at 373 K, converting 25.0 g of ice to water at 273 K while condensing some of the steam to water at 373 K. (a) Find the change in entropy of the ice (b) Find the change in entropy of the steam. (c) Find the change in entropy of the Universe. (30.5 J/K, -22.3 J/K,  8.2 J/K)

... 

Given:

Temperature of ice = `\T_{ice}`  = 273 K

Temperature of Steam =`\T_{steam}` = 373 K

Mass of ice = m = 25 g = 0.025 kg

Latent heat of fusion of water `\H_f`  = 336,000 J/K

To Find:

Change in Entropy of the ice = `\△S_{ice}` = ?

Change in Entropy of the steam = `\△S_{steam}` = ?

Change in Entropy of the universe = `\△S_{universe}` = ?

Solution:

The general formula for change in entropy is 

△S = `\frac {△Q}{T}`

Here △Q is the heat absorbed  from the steam can be calculated as 

△Q = m `\H_f`  

△Q = 0.025 kg 𝐱 336,000 J/Kg

△Q = 8400 J

Calculation of Change in Entropy of the ice:

`\△S_{ice}` = `\frac{△Q}{T_{ice}}`

by putting values

`\△S_{ice}` = `\frac{8400 J}{273 K}`

`\△S_{ice}` = 30.8 JK⁻¹---------------Ans.1

Calculation of Change in Entropy of the steam:

`\△S_{steam}` = -`\frac{△Q}{T_{steam}}`

[As heat will be removed from steam on thermal contact with ice so entropy is negative]

by putting values

`\△S_{steam}` = -`\frac{8400 J}{373 K}`


`\△S_{steam}` = -22.5 JK⁻¹---------------Ans.2

Calculation of Change in Entropy of the universe:

`\△S_{universe}` = `\△S_{steam}` + `\△S_{steam}`

`\△S_{universe}` = 30.8 JK⁻¹ + ( - 22.5  JK⁻¹)

`\△S_{universe}` = 30.8 JK⁻¹ - 22.5  JK⁻¹

`\△S_{universe}` = 8.3 JK⁻¹------------------Ans.