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?
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.
i need help on this problem too
2.3 x 10^-3 kg
around 6 grams of water is needed
To solve this problem, we need to use the equation for heat transfer:
Q = mcΔT
Where:
Q = heat transferred
m = mass
c = specific heat capacity
ΔT = change in temperature
First, we need to calculate the heat transfer for the gold ring. We'll assume the specific heat capacity of gold is 0.129 J/g°C.
Q1 = mcΔT1
Where:
m = mass of the gold ring (47g)
c = specific heat capacity of gold (0.129 J/g°C)
ΔT1 = change in temperature for the gold ring (99°C - 38°C = 61°C)
Q1 = 47g * 0.129 J/g°C * 61°C
Q1 = 377.493 J
Next, we need to calculate the heat transfer for the water. We'll assume the specific heat capacity of water is 4.18 J/g°C.
Q2 = mcΔT2
Where:
m = mass of the water
c = specific heat capacity of water (4.18 J/g°C)
ΔT2 = change in temperature for the water (25°C - 38°C = -13°C)
Now, we can calculate the mass of the water:
Q1 = Q2
47g * 0.129 J/g°C * 61°C = m * 4.18 J/g°C * -13°C
Simplifying the equation:
377.493 J = -54.34m
Divide both sides by -54.34:
m = 377.493 J / -54.34 J/g°C
m ≈ -6.944g
Since mass cannot be negative, we can conclude that there is no mass of water that can cool the gold ring to 38°C from 99°C.
Therefore, no mass of water at 25°C is needed in this scenario.