A steam engine intakes 200g of steam at 98 degree celsius per minute and cools it down to 10 degree celsius. Calculate the heat rejected by the steam engine per minute. Latent heat of vaporization of steam= 540 cal/g.

The quantities like pressure, volume, and temperature which helps us to study the behaviour of a thermodynamic system is called thermodynamic variable.

Internal energy of system contains: kinetic energy and intermolecular potential energy.

Solution:

Given,

height of water (h) = 350 m

mass of water (m) = ?

Let acceleration due to gravity be g = 10 m/s^2

Then, the potential energy of water at height h (PE) = mgh

= m (10) (350)

= (3500 m) Joules

Again, the amount of heat energy gained by water is given by (Q) = ms 'dt' where s is the specific heat capacity of water and 'dt' is the difference in temperature.

Question says that the entire PE is converted to Q so,

PE = Q

or, 3500 m = ms 'dt'

or, 3500 = 4200 'dt'

So, 'dt' = 3500/4200 = 5/6 K

Hence,...

The amount of heat required to change the temperature of 1 mole of gas through 1 K is defined as molar heat capacity of the gas.

From First Law of Thermodynamics,

dQ = dU + dW

At constant volume, dW = 0. But it is not 0 at constant pressure. Hence, more heat has to be supplied to bring the same change in the internal energy of the gas. Therefore, the molar heat capacity at constant pressure is more than that at constant volume.

Solution:

Given,

(dQ)_P = 105 calories

no of moles (n) = 3

initial temperature (T1) = 25 + 273 = 298 K

final temperature (T2) = 30 + 273 = 303 K

change in temperature (dT) = 303 - 298 = 15 K

We know,

Cp = (dQ)_P / n'dT'

= (105 * 4.2) / (3 * 15)

= (9.8)

And,

Cp - Cv = R

or, (9.8) - (8.31) = Cv

So, Cv = 1.5

Now,

Let (dQ)v be the amount of heat required to raise the temperature of the gas through same difference at constant difference then,

(dQ)v = nCvdT

= (3 * 1.5 * 15)

= 67.5 J

= 16.07 cal

A thermodynamic process in which the temperature of the system remains constant is defined as isothermal process.

Work done by Gas from P1 to P2 at Constant Temperature

Solution:

n = 5

P1 = 1 atm

V1 = 50 liter

V2 = 5 liter

T = 20oC = (273 + 20) K = 293 K

At constant temperature,

Work done = nRT ln(V2/V1)

= (5 * 8.314 * 293) ln(5/50)

= - 28 kJ

The negative sign indicates that the...

1. The variations in enthalpy that can not be detected per calorimeter can be detected with the aid of

1. Newton’s law


2. Hess’s law


3. Krebs law


4. Ohm’s law

2. The energy required to sever a given covalent bond is named

1. bond energy


2. bond enthalpy


3. bond dissociation energy


4. all of above

3. Changes in enthalpy in an exothermic reaction is

1. positive


2. negative


3. constant


4. neutral

4. The first law of thermodynamics states that energy can not be

1. created only


2. destroyed only


3. converted


4. created and destroyed

5. Hess...

1. Which of the following is correct expression of the first law of thermodynamics?

a) q=H - W

b)H=q+W

c)U= q+ W

d)U=H +PV

2.The first law of thermodynamics fails to decide____________

a) the direction of the the process

b) the extent of conversion of one form of energy to another

c) both these

d) none of these

4.The first law of thermodynamics is also known as,

a) law of mass action

b) law of conservation of mass

c) law of conservation of energy

d) law of conservation of mass and energy

5. Heat cannot by...

From the ideal gas equation, the relation pressureP, volumeV, and temperature of the gas can be expressed as,

…… (1)

Here,nis number of moles.

From the above formula, the temperature of the gas at pointAin the figure P 12.67 can be taken as,

SubstituteforP,forV, 10 mol forn, and 8.31 forR.

Therefore, the temperature of the gas at point A is.