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Physics Thermodynamics

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2 months ago

Physics Thermodynamics Class 11th

A Block of ice of mass 120g at temperature $0 ° C$ is put in 300g of water at $25 ° C .$ The xg of ice melts as the temperature of the water reaches $0 ° C .$ The value of x is _________.

[Use specific heat capacity of water = 4200Jkg^-1K^-1, Latent heat of ice = 3.5 * 10⁵Jkg^-1]

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Payal Gupta

Contributor-Level 10

Using concept of calorimeter, we can write

Gain in heat = Loss in heat

$x * L = m c Δ T \Rightarrow x * 3.5 * 1 0^{5} = \frac{300}{1000} * 25 * 4200$

$\Rightarrow x = \frac{300 * 25 * 4200}{1000 * 3.5 * 1 0^{5}} = \frac{3 * 25 * 42}{35 * 1000} = \frac{90}{1000} k g = 90 g m$

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2 months ago

Physics Thermodynamics Thermodynamics

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are -2220.0 kJ mol^-1, -393.5kJ mol^-1 and -285.8kJ mol^-1 respectively. The magnitude of the enthalpy of formation of propane C₃H₈ is _________ kJ mol^-1 (Nearest integer)

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Vishal Baghel

Contributor-Level 10

3C (Graphite) + 4H₂ (g) -> C₃H₈ (g)

$Δ H_{f} (C_{3} H_{8}) = [3 * Δ H_{c o m b} (C)] + [4 * Δ H_{c o m b} (H_{2})] - [Δ H_{c o m b} (C_{3} H_{8})]$

= -10.3.7 kJ/mole

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2 months ago

Physics Thermodynamics Thermodynamics

The pressure P₁ and density d₁ of diatomic gas $(γ = \frac{7}{5})$ changes suddenly to P₂(>P₁) and d₂ respectively during an adiabatic process. The temperature of the gas increases and becomes ________times of its initial temperature. (given $\frac{d_{2}}{d_{1}} = 32)$

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Vishal Baghel

Contributor-Level 10

For adiabatic process – PV^Y= const

$T_{1} {V_{1}}^{Y - 1} = T_{2} V_{2}^{Y - 1}$

$\frac{T_{2}}{T_{1}} = {(\frac{v_{1}}{v_{2}})}^{Y - 1} = {(\frac{d_{2}}{d_{1}})}^{Y - 1} = {(32)}^{(\frac{7}{5} - 1)} = {(32)}^{2 / 5}$

$= {(2)}^{2} = 4$

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2 months ago

Physics Thermodynamics Class 11th

A steam engine intakes 50g of steam at $100 ° C$ per minute and cools it down to $20 ° C .$ If latent heat of vaporization of steam is 540 cal g^-¹, then the heat rejected by the steam engine per minute is………….* 10³ cal.

(Given : specific heat capacity of water : 1 cal g^-¹ ⁰C^-¹)

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Payal Gupta

Contributor-Level 10

$Q_{1} = \underset{w a t e r}{20 ° C} \to 100 ° s t e a m$

$Q_{1} = m c Δ T + m L$

$= m [c Δ T + L]$

$= 31000 = 31 * 1 0^{3} c a l$

$Q_{2} = \frac{293}{373} * 31 * 1 0^{3}$

$= 24.35 * 1 0^{3} c a l$

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2 months ago

Physics Thermodynamics Class 11th

In an experiment to verigy Newton's law of cooling, a graph is plotted between, the temperature difference $(Δ T)$ of the water and surrounding and time as shown in figure. The initial temperature of water is taken as $80 ° C .$ The value of t₂ as mentioned in the graph will be _______.

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Payal Gupta

Contributor-Level 10

$\frac{d T}{d t} = - k (T - T_{0})$

$\frac{d T}{(T - T_{0})} = - k d t$

$\Rightarrow k = - \frac{1}{6} l n (\frac{2}{3})$ - (1)

Now $\frac{d T}{d t} = - k (T - T_{0})$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

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2 months ago

Physics Thermodynamics Class 11th

A monoatomic gas performs a work of $\frac{Q}{4}$ where Q is the heat supplied to it. The molar heat capacity of the gas will be ________ R during this transformation. Where R is the gas constant.

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Payal Gupta

Contributor-Level 10

$Q = Δ υ + w \Rightarrow n C Δ T = n C_{v} Δ T = \frac{n C Δ T}{4} = \frac{3}{4} n C Δ T = n C_{v} Δ T$

$C = \frac{4}{3} C_{v} = \frac{4}{3} * \frac{3}{2} R = 2 R$

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2 months ago

Physics Thermodynamics Class 11th

A 100g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60ms^-1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?

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Payal Gupta

Contributor-Level 10

Heating energy $= \frac{1}{4} t h$ energy of the hammer

$= \frac{1}{4} * \frac{1}{2} m v^{2}$

$= \frac{1}{4} * \frac{1}{2} * 1.5 * 60 * 60$

$= \frac{15 * 6 * 60}{4 * 2}$

= 675 J

$Δ T = \frac{675}{42} = 16.07 ° C$

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2 months ago

Physics Thermodynamics Thermodynamics

A cornot engine whose heat sinks at 27°C has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

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Vishal Baghel

Contributor-Level 10

0.25 = 1 - $\frac{T_{2}}{T_{1}}$ $\frac{_{}}{_{}}$

0.25 = 1 - $\frac{(27 + 273)}{T_{1}} \Rightarrow \frac{300}{T_{1}} = 1 - 0.25$ $\frac{}{_{}} \frac{}{_{}}$

$\Rightarrow \frac{300}{T_{1}} = 0.75$

$T_{1} = \frac{300 * 100}{75}$

T1 = 400 k

Now, efficiency increases by 100%

$η_{2} = 100 % η_{1} + η_{1}$

$= 2 η_{1}$

= 0.25 * 2

= 0.50

0.50 = 1 - $\frac{T_{2}'}{T_{1}}$ $\frac{_{}}{_{}}$

$\Rightarrow \frac{T_{2}'}{T_{1}} = 0.50$

$T_{1}' = \frac{300}{0.5}$

$T_{1}' = 600 k$

$T_{1}' - T_{1} =$ (600 – 400) = 200 k or 200° C

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2 months ago

Physics Thermodynamics Class 11th

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

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alok kumar singh

Contributor-Level 10

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

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2 months ago

Physics Thermodynamics Mechanical Engineering

Which is the best Mechanical Engineering book to understand the basics of thermodynamics for beginners?

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Vidhi Jain

Contributor-Level 10

One of the most important Mechanical Engineering subjects is Thermodynamics, which deals with the concepts of heat energy and temperature, along with its conversion between different forms. You would come across a variety of books that will give you a basic understanding of the concept, but here are a few book suggestions for Mechanical Engineering freshers to grasp the basics of thermodynamics quickly:

Basic and Applied Thermodynamics by P. K. Nag
Engineering Thermodynamics by S. K. Som & G. Biswas
The Laws of Thermodynamics: A Very Short Introduction by Peter Atkins

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