Question

You use a 15.0 gram piece of aluminum foil to cover a pan in the oven. The specific heat for aluminum is c = 0.900 J/g o C. If the temperature is raised from 25 o C to 350 o C, how much heat was absorbed?

Answer 1

Answer:

Best regards.

Explanation:

Hello,

In this case, we relate the heat, mass, heat capacity and temperature when a thermal change is carried out as shown below:

$Q=mCp(T_{final}-T_{initial})$

Now, for the given data, we compute the absorbed heat (due to the temperature increase) as follows:

$Q=15.0g*0.900\frac{J}{g^oC}*(350^oC-25^oC) \\\\Q=4.39x10^3J=4.39kJ$

Best regards.

Answer 2

Answer:

4,387.5 J was absorbed

Explanation:

Calorimetry is the part of physics that is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.

In this way, there is a direct proportional relationship between heat and temperature. Thus, the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous) is calculated using the following expression:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

• Q= ?
• c= 0.900 $\frac{J}{g*C}$

• m= 15 g
• ΔT=Tfinal - Tinicial= 350 °C - 25 °C= 325 °C

Replacing:

Q= 0.900 $\frac{J}{g*C}$ *15 g * 325 °C

Q=4,387.5 J

4,387.5 J was absorbed