A 12-inch diameter ball of pure cobalt metal at 225 °C was placed in 10.0 gal.of water at 15.0 °C.What is the final temperature of the water? Assume no heat is lost to the surroundings.(specific heat for cobalt is 0.421 J/g °C,density of cobalt = 8.862 g/cm³,1 gal.= 3.785 L,1 in = 2.54 cm,V(sphere)= 4/3 πr³)

Question

Quizplus · Accepted Answer

First, we need to find the mass of the cobalt ball using its diameter (12 inches) and density (8.862 g/cm³):
radius (r) = 12 / 2 = 6 inches = 15.24 cm
V(sphere) = (4/3)πr³ = (4/3)π(15.24)³ = 14614.5 cm³
mass = volume x density = 14614.5 cm³ x 8.862 g/cm³ = 129486 g

Next, we can use the formula for heat transfer:
q = m x c x ∆T
where q is the heat transferred, m is the mass of the cobalt, c is the specific heat of cobalt, and ∆T is the change in temperature.

Assuming no heat is lost to the surroundings, we can say that the heat lost by the cobalt is equal to the heat gained by the water:
q(cobalt) = -q(water)

The heat lost by the cobalt can be calculated as:
q(cobalt) = m x c x ∆T
where m is the mass of the cobalt, c is the specific heat of cobalt, and ∆T is the change in temperature.

The mass of the water can be found using its volume:
10.0 gal. = 10.0 x 3.785 L/gal. = 37.85 L = 37.85 kg

The heat gained by the water can be calculated as:
q(water) = m x c x ∆T
where m is the mass of the water, c is the specific heat of water (4.184 J/g°C), and ∆T is the change in temperature.

Now we can set up the equation:
m(cobalt) x c(cobalt) x ∆T(cobalt) = -m(water) x c(water) x ∆T(water)
129486 g x 0.421 J/g°C x (225 - ∆T(cobalt)) = -37.85 kg x 4.184 J/g°C x (∆T(water) - 15.0°C)

Simplifying and solving for ∆T(water):
∆T(water) = (129486 g x 0.421 J/g°C x (225 - ∆T(cobalt))) / (37.85 kg x 4.184 J/g°C) + 15.0°C

After simplification, we get:
∆T(water) = 54.5 - 0.0025 ∆T(cobalt)

We can solve for ∆T(cobalt) by setting the heat lost by the cobalt equal to the heat gained by the water:
m(cobalt) x c(cobalt) x ∆T(cobalt) = -m(water) x c(water) x ∆T(water)
129486 g x 0.421 J/g°C x (∆T(cobalt) - 225) = -37.85 kg x 4.184 J/g°C x (∆T(cobalt) - 15.0°C)

Simplifying and solving for ∆T(cobalt):
∆T(cobalt) = (37.85 kg x 4.184 J/g°C x 15.0°C + 129486 g x 0.421 J/g°C x 225) / (129486 g x 0.421 J/g°C + 37.85 kg x 4.184 J/g°C)

After simplification, we get:
∆T(cobalt) = 154.5°C

Therefore, the final temperature of the water is:
final temperature = 54.5 - 0.0025 x 154.5 = 69.5°C

Therefore, the best choice is C, which is the final temperature of the water.

A 12-Inch Diameter Ball of Pure Cobalt Metal at 225