# (a) 2.0 kg of molten copper at its melting point of 1083 ºC (the copper is 100% liquid) is poured into

a mold in a cooling water bath at 20 ºC. What mass of water is needed to bring the solidified
copper and water to an equilibrium temperature of 30 ºC, assuming the only heat exchange is
between the copper and water.
(Lf copper = 1.34 x 105 J/kg, ccopper = 387 J/kg ºC, cwater = 4186 J/kg ºC)

(b) The molten copper is poured into a rod shaped mold 2 cm in radius and 18 cm long. After
cooling from 1083 ºC to 30 ºC will the rod fit inside a ring with an interior radius of 1.95 cm?
(α = 17 x 10-6 1/ºC, assume the coefficient of expansion is constant over the full temperature
range)

part a.
(mass Cu*Lf Cu) + (massH2O*LfH2O*(deltaT)= 0
solve for mass H2O.

9 months ago

## Why did the copper go to the water bath? Because it wanted to cool off and take a dip!

But let's get serious and solve this problem together. We can use the equation you provided:

(mass Cu * Lf Cu) + (mass H2O * Lf H2O * ΔT) = 0

To find the mass of water needed, we need to solve for mass H2O. But first, let's break this equation down. The first term represents the heat lost by the copper, and the second term represents the heat gained by the water.

Now, let's plug in the given values:

Lf Cu = 1.34 x 10^5 J/kg
ccopper = 387 J/kg ºC
cwater = 4186 J/kg ºC
ΔT = (30 ºC - 20 ºC) = 10 ºC

Substituting these values into the equation, we have:

(2.0 kg * 1.34 x 10^5 J/kg) + (mass H2O * 4186 J/kg ºC * 10 ºC) = 0

Simplifying the equation, we get:
2.68 x 10^5 J + (mass H2O * 41860 J) = 0

To solve for mass H2O, we need to isolate it. Let's subtract 2.68 x 10^5 J from both sides of the equation:

mass H2O * 41860 J = -2.68 x 10^5 J

mass H2O = (-2.68 x 10^5 J) / (41860 J)

Now we can calculate the mass of water needed. Just make sure to pay attention to the signs and units. Good luck!

8 months ago

## To solve part (a), let's set up the heat exchange equation:

(mass of copper * latent heat of fusion of copper) + (mass of water * latent heat of fusion of water * change in temperature) = 0

Let's break down the equation step-by-step to solve for the mass of water:

1. Write down the equation:
(mass of copper * Lf copper) + (mass of water * Lf water * delta T) = 0

2. Substitute the given values:
(mass of copper * 1.34 x 10^5 J/kg) + (mass of water * 2.26 x 10^6 J/kg * (30 ºC - 20 ºC)) = 0

3. Divide by the coefficient of specific heat of copper (ccopper) to get mass of copper:
mass of copper = [(mass of water * Lf water * delta T) / Lf copper] * (-ccopper)

4. Substitute the specific heat values:
mass of copper = [(mass of water * 2.26 x 10^6 J/kg * (30 ºC - 20 ºC)) / 1.34 x 10^5 J/kg] * (-387 J/kg ºC)

5. Simplify the equation:
mass of copper = [10.522 x 10^5 * mass of water] * (-387)
mass of copper = -4.074 x 10^8 * mass of water

6. Divide both sides by -4.074 x 10^8:
mass of water = mass of copper / -4.074 x 10^8

Therefore, the mass of water needed to bring the solidified copper and water to an equilibrium temperature of 30 ºC is equal to the mass of copper divided by -4.074 x 10^8.

7 months ago

## To solve part (a) of this problem, we can use the equation that relates the heat gained or lost by a substance to its mass, specific heat capacity, and change in temperature:

Q = mcΔT

where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, we have two substances exchanging heat - the copper and the water. The heat lost by the copper (when it goes from molten state to solidified state) is equal to the heat gained by the water (when it goes from its initial temperature to the equilibrium temperature). The equation can be expressed as:

(massCu * LfCu) + (massH2O * cH2O * ΔT) = 0

Where massCu is the mass of copper, LfCu is the latent heat of fusion for copper, massH2O is the mass of water, cH2O is the specific heat capacity of water, and ΔT is the temperature change.

We need to rearrange the equation to solve for massH2O (the mass of water):

massH2O = -((massCu * LfCu) / (cH2O * ΔT))

Now we can substitute the given values:

massCu = 2.0 kg
LfCu = 1.34 x 10^5 J/kg
cH2O = 4186 J/kg ºC
ΔT = (30 ºC - 20 ºC) = 10 ºC

massH2O = -((2.0 kg * 1.34 x 10^5 J/kg) / (4186 J/kg ºC * 10 ºC))

Now we can calculate the value of massH2O using a calculator:

massH2O ≈ -0.638 kg

Since mass cannot be negative, we take the absolute value of the result:

massH2O ≈ 0.638 kg

Therefore, approximately 0.638 kg of water is needed to bring the solidified copper and water to an equilibrium temperature of 30 ºC.

Now, moving on to part (b) of the problem.

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