Question

A system absorbs 6kj of heat and does 1.5 kj of work on it's surroundings. The change in internal energy is________ a) - 7.5 kj b) - 4.5 kj c) +4.5kj d) +7.5 kj

Answer 1

ΔQ + ΔW = ΔU

since the heat is absorbed by the system ΔQ is positive.

And work is done on the surroundings ΔW is negative.

6 - 1.5 = ΔU

ΔU = 4.5 kJ

Answer 2

The change in internal energy is +4.5 kJ

Explanation:

According to first law of thermodynamics:

$\Delta E=q+w$

$\Delta E$=Change in internal energy  = ?

q = heat absorbed or released

w = work done or by the system

w = work done by the system=$-P\Delta V$  {Work done by the system or on the surroundings  is negative as the final volume is greater than initial volume }

w = -1.5 kJ

q = +6 kJ   {Heat absorbed by the system is positive}

$\Delta E=+6kJ+(-1.5)kJ=+4.5kJ$

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