Question

A pure gold ring with a volume of 1.79 cm^3 is initially at 17.4 ∘C. When it is put on, it warms to 28.5 ∘C. How much heat did the ring absorb? (density of gold =19.3 g/cm^3)

Answer 1

Answer:

Hi Adriana!. You should need to find out which is the specific heat for gold. As this value is 0.129 J/g °C.  The ring has absorbed 49.5 J of heat.

Explanation:

First of all, you have to use the density formula to find out your mass.

d = mass/volume

19.3 g/cm3 = mass / 1.79 cm3

19.3 g/cm3 x 1.79 cm3 = mass

34.55 g = mass

Now, that we have the gold mass, let's go to the specific heat formula

Heat = mass . Specific heat . ΔT (where ΔT means the difference between temperatures.- Tfinal - Tinitial) So..

Heat = mass . Specific heat . ΔT

Heat = 34.55 g x 0.129 J/g °C x (28.5°C - 17.4°C)

Heat = 34.55 g x 0.129 J/g °C x (11.1°C)

Heat = 49.5 J