Question #69805

1 Answer
Dwight
Oct 6, 2017

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

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