Answer :
The question is incomplete as the samples are not mentioned in the question. The samples are of the elements Aluminium, brass, copper and platinum.
If 500 j of energy is added to 1 kg of the given samples, Aluminium will experience the least temperature increase.
The relation between heat given(Q) and temperature change(delta) is given by the following equation
Q = mSΔT
Here m is the mass of the given substance.
And S is the specific heat capacity.
In the given case the heat is given and mass of a substance is the same
So, ΔT ∝ 1/S
Therefore the substance with the highest specific heat has the lowest temperature change.
The specific heats of Aluminium, Brass, Copper, and Platinum are 900, 380, 390, and 130 J/kg K respectively.
As the specific heat of Aluminium is the highest among the given materials, the least temperature increase would be in the case of aluminum.
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