How much energy is needed to heat 1 kg of sand, which has a specific heat of 664 J/(kg K), from 30°C to 50°C?

Q = nC∆T

here Q = energy required

n = amount of substance

C = specific heat

then

Q = 1×664×(50-30) = 13280 j Answer

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.