Asked by Demi
A Jeweler working with a heated 47g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initially at 99 degrees Celsius, what mass of water at 25 degrees Celsius is needed to lower the ring's temperature to 38 degrees Celsius?
Answers
Answered by
drwls
You need the specific heat of gold, as well as that of water.
If you dopn't know it, see
http://hyperphysics.phy-astr.gsu.edu/Hbase/tables/sphtt.html
Let's call the specific heats Cg and Cw. The heat loss of the gold in going from 99 to 38 C equals the heat energy gained by the water in going from 25 to 38 C.
0.47g*Cg*61 = M*Cw*13
M = 0.47g * (Cg/Cw) *(61/13)
Gold has a very low specific heat compared to water. You may surprised at how little water it takes to cool it. I was.
If you dopn't know it, see
http://hyperphysics.phy-astr.gsu.edu/Hbase/tables/sphtt.html
Let's call the specific heats Cg and Cw. The heat loss of the gold in going from 99 to 38 C equals the heat energy gained by the water in going from 25 to 38 C.
0.47g*Cg*61 = M*Cw*13
M = 0.47g * (Cg/Cw) *(61/13)
Gold has a very low specific heat compared to water. You may surprised at how little water it takes to cool it. I was.
Answered by
Emily
i need help on this problem too
Answered by
KLD
2.3 x 10^-3 kg
Answered by
Khanh
around 6 grams of water is needed
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