Question

when 1564 J of heat energy is added to a sample of gold at 25.0°C, the temperature of the gold increases to 449°C. what is the means of the gold sample? (specific heat of gold= 0.129 J/g-°C

Answer 1

The mass of the gold sample is 28.59 g

Step-by-step explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system. The amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without changing its physical state (solid, liquid or gaseous) is calculated by:

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= 1564 J

• c= 0.129
  `(J)/(g*C)`

• m=?
• ΔT= Tfinal - Tinitial= 449°C - 25°C= 424 °C

Replacing:

`1564 J=0.129(J)/(g*C)*m*424 C`

Solving:

`m=(1564 J)/(0.129(J)/(g*C)*424 C)`

m=28.59 g

The mass of the gold sample is 28.59 g