Answer to Question #130953 in Molecular Physics | Thermodynamics for Amkit

Question

A 5hp fan is used in a large room to provide for air circulation. Assuming a well insulated sealed room.determine the increase in internal energy after 1 hour of operation

Answer 1

Assuming that $Q=0$ and $\Delta P=\Delta K=0$ , the first law becomes

$Q-W=\Delta U$

$-W=\Delta U$

 The work input is therefore

$W=-5\times1\times746=-3730J$

multiply by 3600 to get $J/s$

$W=J/s$

$3730\times3600=-1.343\times10^7 J$

The negative sign indicates that the work is input to the system

Therefore

$\Delta U=1.343\times10^7J$