1.

What is Delta U for the process described by figure. Heat supplied during the process q= 100kJ

Answer»

`+50kJ`
`-50kJ`
`-150kJ`
`+150kJ`

SOLUTION :`DELTA U = q + W`
W = area under graph`= (1)/(2) xx 1 xx (1 +2) = (3)/(20 xx 10^(5)N - m^(2)`
`Delta U = +100 - (3)/(2) xx (10^(5))/(1000) = 100-150= -50J`


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