210 J of heat is added to a system having an internal energy of 58 J. Calculate the amount of the external work done by the system.
From the first law of thermodynamics
"\\Delta U=Q+W"
"\\Delta U=58J"
"Q=210J"
"W=?"
"\\therefore W=58-210 = -152J"
The amount of external work done is 152J
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