Home
Class 11
PHYSICS
1 cc of water becomes 1681 cc of steam w...

1 cc of water becomes 1681 cc of steam when boiled at a pressure of `10^5 Nm^2`. The increasing internal energy of the system is (L.T. of steam is 540 cal `g ^(-1)`, 1 calorie = 4.2J)

A

300 cal

B

500 cal

C

225 cal

D

600 cal

Text Solution

Verified by Experts

The correct Answer is:
B
Work done (W)= `P(V_2-V_1`)
`=10^5 (1681-1)xx 10^(-6) = 168J =(168)/(4.2)` = 40 cal
heat supplied to the 1g of water
`Delta Q = m L = 1xx540` = 540 cal
Now, `Delta Q=Delta W+Delta U`
`implies Delta U=Delta Q-Delta W=540-40` = 500 cal.
Promotional Banner

Similar Questions

Explore conceptually related problems

1 g of water (volume 1 cm^(3) becomes 1671 cm^(3) of steam when boiled at a pressure of 1atm. The latent heat of vaporisation is 540 cal/g, then the external work done is (1 atm = 1.013 xx 10^(5) N/ m^(2))

One kg of water at 373K is converted into steam at the same temperature. The volume 1 cm^(3) of water becomes 1671 cm^(3) on boiling. Calculate the change in internal energy of the system , if heat of vaporisation is 540 cal g^(-1) . Given standard atmospheric pressure =1.013xx10^(5) Nm^(-2) .

1 g of water at 100^(@)C is completely converted into steam at 100^(@)C . 1g of steam occupies a volume of 1650cc. (Neglect the volume of 1g of water at 100^(@)C ). At the pressure of 10^5N//m^2 , latent heat of steam is 540 cal/g (1 Calorie=4.2 joules). The increase in the internal energy in joules is

1g of water at 100^(@)C is converted into steam at the same temperature. If the volume of steam is 1671 cm^(3) , find the change in the internal energy of the system. Latent of steam =2256 Jg^(-1) . Given 1 atmospheric pressure = 1.013xx10^(5) Nm^(-2)

When water is boiled at 2 atm pressure the latent heat of vapourization is 2.2xx10^(6) J//kg and the boiling point is 120^(@)C At 2 atm pressure 1 kg of water has a volume of 10^(-3)m^(-3) and 1 kg of steam has volume of 0.824m^(3) . The increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120^(@)C is [ 1 atm pressure =1.013xx10^(5)N//m^(2) ]

One gram of water (1 cm^3) becomes 1671 cm^3 of steam when boiled at a constant pressure of 1 atm (1.013xx10^5Pa) . The heat of vaporization at this pressure is L_v=2.256xx10^6J//kg . Compute (a) the work done by the water when it vaporizes and (b) its increase in internal energy.

One gram of water (1 cm^(3)) becomes 1671 cm^(3) of steam at a pressure of 1 atm. The latent heat of vaporization at this pressure is 2256 J/g. Calculate the external work and the increase in internal energy.