Question

The specific heat of aluminum is 0.125 cal/g °C. If 12.5 grams of aluminum were heated from 20.0 °C to 100.0 °C, calories of heat energy would be absorbed by the aluminum.

Answer 1

Answer:

$Q=125cal$

Explanation:

Hello,

In this case, this could be determined via the following equation for the determination of the absorbed heat by the aluminium, considering the heat capacity, the change in the temperature and the heated mass:

$Q=mCp\Delta T\\Q=mCp(T_2-T_1)\\Q=12.5g*0.125\frac{cal}{g*^oC}*(100.0-20.0)^oC\\ Q=125cal$

Best regards.

Answer 2

Aluminum does not undergo any phase change from 20.0°C to 100.0°C, therefore we only consider sensible heat change, which is calculate as
q = mc(dT)q = (12.5 g)(0.125 cal/g-°C)(100 - 20)°C = 125 cal

Therefore, 125 calories of heat are absorbed by the aluminum.