100g of water is heated from 30oC to 50oC. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184J/Kg/K)

ΔQ=ΔU+ΔW
Since the volume remains constant (ignoring slight expansion), ΔW = 0.
Therefore, ΔQ = ΔU.
The heat absorbed by the water is given by:
ΔQ = mcΔT
where:
m = mass of water = 100g = 0.1kg
c = specific heat capacity of water = 4184 J/kg/K
ΔT = change in temperature = 50oC - 30oC = 20oC = 20K
ΔQ = (0.1 kg)(4184 J/kg/K)(20 K) = 8368 J = 8.368 kJ
Approximating to the nearest option, the change in internal energy is 8.4 kJ.