When a 120 g sample of aluminum absorbs 9612 of heat energy, its temperature increases from 25°c to 115°c. find the specific heat of aluminum

0.89 J/g°C

Concept tested: Quantity of heat

We are given;

                                                      = 90°C

We are required to calculate the specific heat capacity;

That is;

Q = m × c × ΔT

Specific heat capacity, c = Q ÷ mΔT

                                         = 9612 J ÷ (120 g × 90°C)

                                         = 0.89 J/g°C

Therefore, the specific heat capacity of Aluminium is  0.89 J/g°C

the volume increases

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