A steam engine intakes 100 g of steam at `100^(o)C` per minute and cools it down to `20^(o)C`. Calculate the heat rejected by the steam engine per minute. Latent heat of vaporization of steam `=540 cal g^(-1)`.

Answer

A, C

Explanation

`A=25cm^(2)=25 XX10^_(-4)m^(2)`
`L=1 mm =10^(-3)m,K=50W//m^(@)C.`
(Q)/(T)=rate of conversion of water in to steam
`Q/T=(100xx10^(-3)xx2.26xx10^(6))/(1 min)`
`=(100xx10^(3)xx2.26)/(1 min)`
`2.26/6 xx10^(4)`
`=0.376xx10^(4) J//s`
`(Q)/(T)=(kA(T_1-T_2))/(l)`
`implies 0.376xx10^(4)=(50xx25xx10^(-4)xx(T-100))/(10^(-3))`
`implies(T-100)=(10^(-3)xx0.376xx10^(4))/(50xx25xx10^(-4))`
`=(10^(-3)xx0.376)/(50xx25)`
`implies(T-100)=3.008xx10^(-4)xx10xx(-3)xx10^(8)`
`implies(T-100)=3.008xx10`
`=30^(@)C=30`
`:. T=100+30=130^(@)C.`