InterviewSolution
Saved Bookmarks
| 1. |
The volume of steam produced by `1g` of water at `100^(@)C` is `1650 cm^(3)`. Calculate the change in internal energy during the change of state. Given `J= 4.2xx10^(7) erg`. `cal.^(-1), d= 981 cm^(s-2)`. Latent heat of stream `= 540 cal. G^(-1)` |
|
Answer» Here, mass of water `= 1g` `:.` Initial volume of water, `V_(1)= 1cm^(3)` Volume of stream, `V_(2)= 1650 cm^(3)` change of internal energy, `dU=?` As the state of water is changing, `:. dQ=mL` `= 1xx540 cals= 540xx4.2xx10^(7) ergs` `22.68xx10^(9) ergs` Taking `P= 1` atmosphere `= 76xx13.6xx981 dyn e cm^(-2)` `dW=P dV= P(V_(2)-V_(1))` `76xx13.6xx981(1650-1)` `=76xx13.6xx981xx1649 erg` As `dQ=dU+dW` `:. dU=dQ-dW` `= 22.68xx10^(9)-1.67xx10^(9)` `= 21.01xx10^(9) ergs` |
|