Question

Question #69805

Answer 1

This heat effect would be seen by sing 2.98 g of Al.

Explanation:

This solution of this problem will require that you look up the specific heat, `C`, of aluminum, as it is required in the formula that relates the quantities of mass, temperature change and energy:

`E=mxxCxxDeltat`

where `Deltat` is the temperature change (in Kelvin or Celsius) and `m` is the mass of the sample.

To save you the trouble, the specific heat of Al is 0.900 J/g °C (meaning that it takes 0.900 J of energy to warm 1.0 g of Al by 1.0 °C)

Putting it all together:

`150.0J=mxx0.900xx56°C`

`m=150.0/(0.900xx56) = 2.98` g