Question

In an internal combustion engine, during the compression strokes, the heat rejected to the cooling water is 75 J/g and the work input is 120 J/g. The change in internal energy of the working fluid is:

• A. -45 J/g
• B. +45 J/g
• C. -195 J/g
• D. +195 J/g

Hint:

First law of thermodynamics: \( \Delta U = Q + W \) (signs must be carefully considered).

Answer 1

The Correct Option is B

Step 1: Apply the first law of thermodynamics.
\[\Delta U = Q + W\]where \( Q \) denotes heat interaction and \( W \) denotes work input.

Step 2: Substitute values.
Given \( Q = -75 \, \text{J/g} \) (heat rejected) and \( W = +120 \, \text{J/g} \). \[\Delta U = -75 + 120 = +45 \, \text{J/g}.\]

Final Answer: \[\boxed{\text{B) +45 J/g}}\]