**Solution (a):** Heat absorbed, ΔQ = 1176 J Work done, ΔW = 352.8 J

Change in internal energy, Δu =?

Apply first law of thermodynamics, ΔQ = Δu + ΔW, we have, 1176 = Δu + 352.8

⇒ Δu = 1176 – 352.8 ⇒ Δu = **823.2 J**

**Solution (b): **Heat absorbed, ΔQ = 1050 J Work done, ΔW = 84 J

Change in internal energy, Δu =?

Apply first law of thermodynamics, ΔQ = Δu + ΔW, we have, 1050 = Δu + 84

⇒ Δu = 1050 – 84 ⇒ Δu = **966 J**

**Solution (c): **Heat removed, ΔQ = -210 J Work done, ΔW = 0 (because volume is constant)

Change in internal energy, Δu =?

Apply first law of thermodynamics, ΔQ = Δu + ΔW, we have, -210 = 0 + Δu ⇒ Δu =** -210 J**