Physics Thermodynamics

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5 * 10³ J?

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Contributor-Level 10

$\frac{Δ U}{t} = \frac{6000}{60} - 90$

= 10 J

$t = \frac{Δ U}{10} = \frac{2.5 * 1 0^{3}}{10} = 2.5 * 1 0^{2}$ second

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2 months ago

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

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Payal Gupta

Contributor-Level 10

W_total = W_D→E + W_E→F = Area of triangle DEF

= $\frac{1}{2} * 3 * 300 = 450 J .$

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2 months ago

The ratio of species heats $(\frac{C_{P}}{C_{V}})$ in terms of degree of freedom (f) is given by:

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alok kumar singh

Contributor-Level 10

$\frac{C_{P}}{C_{v}} = y = 1 + 2 / f$

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2 months ago

In a carnot engine, the temperature of reservoir is 527°C and that of sink is 200 K. If the work done by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is___________* 10⁶ J.

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Contributor-Level 10

T₂ = 200 k

$T_{1} = 527 ° C + 273 = 800 K$

w = 12000 kJ = 12 * 10⁶ J

Q₁ =?

$x = 1 - \frac{T_{2}}{T_{1}} = \frac{w}{Q_{1}} = 1 - \frac{200}{800} = \frac{12 * 1 0^{6}}{Q_{1}}$

$\Rightarrow \frac{3}{4} = \frac{12 * 1 0^{6}}{Q_{1}}$

$\Rightarrow Q_{1} = 16 * 1 0^{6} J$

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2 months ago

A mixture of hydrogen and oxygen has volume 2000 cm³, temperature 300K, pressure 100 kPa and mass 0.76g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:

[Take gas constant R = 8.3 JK^-¹ mol^-¹]

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Contributor-Level 10

$V_{T} = 2000 c m^{3} = 2000 * 1 0^{- 6} m^{3} (T o t a l v o l u m e o f m i x t u r e)$

T = 300 K

$_{}^{}^{}$ $P_{T} = 100 k P a = 100 * 1 0^{3} P a = 1 0^{5}$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T} V = (x_{1} + x_{2}) R T$

$\Rightarrow 1 0^{5} * 200 * 1 0^{- 6} = (x_{1} + x_{2}) * 8.3 * 300$

$\frac{x_{1}}{x_{2}} = \frac{0.06}{0.02} = 3 : 1$

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2 months ago

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, $T_{1}$ and $T_{2}$ . The temperature of the hot reservoir of the first engine is $T_{1}$ . and the temperature of the cold reservoir of the second engine is $T_{2}$ . $T$ is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$ , if both the engines perform equal amount of work?

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alok kumar singh

Contributor-Level 10

W = Q_{1} - Q_{2} = Q_{2} - Q_{3}

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

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

$2 = \frac{T_{1}}{T} + \frac{T_{2}}{T}$

$T = \frac{T_{1} + T_{2}}{2}$

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2 months ago

A certain amount of gas of volume V at $27 ° C$ temperature and pressure $2 * 1 0^{7} N m^{- 2}$ expands isothermally until its volume gets doubled. Late it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be

(Use $γ$ = 1.5):

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Contributor-Level 10

$P_{1} = 2 * 1 0^{7} P a$

$P_{1} v_{1} = P_{2} v_{2}$

Since v₂ = 2v₁ Hence $P_{2} = \frac{P_{1}}{2} (I s o t h e r m a l E x p a n s i o n)$

$P_{2} = 1 * 1 0^{7} P a$

$P_{2} {(v_{2})}^{y} = P_{3} {(2 v_{2})}^{y}$

$P_{3} = \frac{1 * 1 0^{7}}{2^{1.5}} = 3.536 * 1 0^{6} P a$

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2 months ago

Let $η_{1}$ is the efficiency of an engine at T₁ = $447 ° C$ and T₂ = $147 ° C$ while $η_{2}$ is the efficiency at T₁ = $947 ° C$ and T₂ $47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be

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Payal Gupta

Contributor-Level 10

As we know that

$η = 1 - \frac{T_{2}}{T_{1}}, s o$

$\frac{n_{1}}{n_{2}} = \frac{0.42}{0.74} \approx 0.56$

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2 months ago

A Carnot engine operates between two reservoirs of temperatures $900 K$ and $300 K$ . The engine performs $1200 J$ of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is

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Contributor-Level 10

$n = \frac{W}{Q_{h}} = 1 - \frac{300}{900} = \frac{2}{3}$

$Q_{h} = \frac{3}{2} W = 1800 J$

$Q_{L} = O_{h} - W = 600 J$

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2 months ago