Question

A sample of octane burns releasing 2290 J of heat to the surroundings, and the gases produced expands against a piston to do 560 joules of work. Calculate the internal energy change for this reaction.
1- –2850 J
2- –1730 J
3- –2290 J
4- +1730 J
5- +2850 J

Answer 1

Answer: The internal energy change for the reaction is -2850 J

Explanation:

• Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

• Sign convention of work:

Work done for expansion process is taken as negative and work done for compression is taken as positive.

According to the First law of thermodynamics,

$\Delta U=q+w$

where,

$\Delta U$ = internal energy

q = heat absorbed or released  = -2290 J

w = work done = -560 J

Putting values in above equation, we get:

$\Delta U=(-2290)+(-560)\\\\\Delta U=-2850J$

Hence, the internal energy change for the reaction is -2850 J