First Law of Thermodynamics (Chemistry)

Question

Please include a step-by-step solution. Thank you.

1. Calculate the change in the internal energy of a system if 50 kJ of heat is added to it and if the system performs 20 000 J of work.
2. Determine the change in the internal energy of a system if 5.5 x 107 kJ of heat is removed from it and if 2.5 x 107 J of work was done on it.

Answer 1

Internal energy is denoted as ∆E or ∆U

Internal energy ∆U = q + w where q is heat and w is work (this is just something you need to memorize)

1). ∆U = ?

q = +50 kJ (heat added so q is positive)

w = -20,000 J = -20 kJ (work is done BY the system so w is negative)

∆U = 50 kJ + -20 kJ

∆U = 30 kJ

2). ∆U = q + w

q = -5.5x10⁷ kJ (negative because heat is given off to the surrounding; exothermic)

w = 2.5x10⁷ J (positive because work is done ON the system)

At this point I'm not sure if you have a typo and maybe the units of both should be J or kJ, but I see heat is in kJ and work is in J. Whatever the correct units, just calculate q + w to get ∆U