A chef uses a large aluminum spoon with a mass of 38.0 g to stir a pot of boiling water. As a result, the spoon’s temperature increases by 3.60°C. Consider that the specific heat capacity of aluminum is 0.90 J/g–°C. How much heat has the spoon absorbed? For credit, you must show all steps of your calculations.

Respuesta :

Answer:

12,312 J or 0.12312 kJ

Explanation:

q = mCT

q = J of heat  

m = mass

C = heat capacity

T = temp change

q = 38 x 0.90 x 3.6

q = 12312 J

q = 0.12312 kJ