Question

0.1 kg of water at 100 degree Celsius is converted in to steam at the same temperature under one atmospheric pressure.If the volume of steam produced by 1kg of water at 100 degree Celsius is 1671×10 to the power -3 meter cube calculate the change in internal energy in the process latent heat of steam is equal to 2.268 ×10 to the power 6 joule per kg density of water is equal to 1000 kg per meter cube​

Answer 1

Answer:

Mass =10g=0.01kg

P=10  

5

Pa

dQ=Q  

H  

2

​  

O

​  

0  

0

−100  

0

+Q  

H  

2

​  

O

​  

−stem

=0.01×4200×100+0.01×0.5×10  

8

=4200+25000=29200

dW=P×ΔV

Δ=  

0.6

0.01

​  

−  

1000

0.01

​  

=0.01699

dW=PΔV=−0.01699×10  

5

1699J

Explanation: