Thermal physics

If any material is denser there's more mass for the same volume and that will affect the amount of energy needed to change the temperature. Specific heat capacity is the energy needed to raise the temperature of 1kg of the material by 1 degree C.

Solving the question:

Q = VIt
Q = mcΔθ
Therefore
VIt = mcΔθ
p = m/v so m = pv
Therefore
VIt = pvcΔθ
Bring all constants over to one side and all variables over to the other
VI/vΔθ = pc/t
The voltage of the heating element, the current in the heating element, the volume of the liquid and the change in temperature (10 degrees) are the same for water and oil.
The only things that change are p (density), c(specific heat capacity) and t (time to heat)
Therefore
p_wc_w/t = p_oc_o/t_o
(w means water, o means oil)
c_w = 2.5c_o
p_w2.5c_o/t = p_oc_o/t_o
p_w = 1.1p_o
1.1p_o2.5c_o/t = p_oc_o/t_o
Cancel p_o and c_o
1.1*2.5/t = 1/t_o
1.1*2.5 = 2.75
2.75/t = 1/t_o
2.75t_o = t
2.75t_o = t
We want t_o (time for oil)
t_{o =} t * 1/2.75
1/2.75 ~ 0.36
t_o = 0.36t