Ask Question
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)

+1
Answers (1)
  1. 6 June, 08:06
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers