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)

Question

Niko Todd

Chemistry

6 June, 07:52

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)

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Answers (1)

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Jaidyn Frederick · Answer 1

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