When a student chemist transferred the metal to the calorimeter, some water splashed out of the calorimeter. will this technique error result in the specific heat of the metal being reported too high or too low?

It will be reported too low.

Explanation:

To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.

Qmetal + Qwater = 0

Qmetal = -Qwater

The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:

m_metal*s*ΔT_metal = - m_water *c*ΔT_water

s = -m_water *c*ΔT_water / m_metal*ΔT_metal

So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.

it is the most commonly accepted one

it takes on an electronical energy