1. Scenario: The high precision penny-ice problem keep 5 sigfigs in your calcula

Question

1. Scenario: The high precision penny-ice problem keep 5 sigfigs in your calculations. Consider a pre-1982 penny (all copper so we can easily figure out heat transfers without that zinc messing us up) with a radius of 0.9525 cm, a thickness of 0.155 cm, and a mass of 3.11 g. Suppose you raise the temperature up to the melting point of copper and add enough heat so it is basically all melted except for an infinitesimally thin shell of solid copper around it keeping it in the shape of the penny. (I basically want to you pretend it is a molten penny, but retains the shape of the penny until it cools.) The penny is placed onto a cube of ice with sides of 10 cm and a temperature of 0C. Assume the penny remains horizontal.

Draw a schematic diagram of the problem. Decide on variable letters for the important quantiles in the problem like penny radius, thickness, mass, ect... Label them on the diagram. • Write a couple of sentence description or bulleted list describing what happens in the problem. Make sure to include what physical processes are involved.

Questions: a) What mass of ice melts before the penny reaches the temperature of the ice? b) What is the volume of the ice that melts? c) What is the volume of the water that results from the ice that melted? d) How far does the penny sink (what is the distance of the exact center of the penny from the top edge of the ice)? Extra Credit: (5pts to the first team with the correct answer): Determine if water will spill out of the ice cavity. If it yes, calculate the volume of water that spills. If no, calculate the distance of the water level below the top ice surface.

Explanation / Answer

melting point of copper = 10850 C Latent heat of fusion of copper (L) = 205 J/g

Specific heat capacity of copper (C) = 0.385 J/g0C Latent heat of fusion of ice (L0) = 334 J/g

Density of ice at 00C = 0.917 g/ml

Given data: radius of penny = 0.9525 cm thickness of penny = 0.155 cm mass of penny (m) = 3.11 g

Volume of penny (V) = 3.14*(0.9525)^2*0.155 = 0.4417 cm^3 = 0.4417 ml

Now heat given off by copper in cooling = m*L + m*C*(1085-0) = 1936.67 J

Now using the Zeroth law of thermodynamics; Heat given by copper = heat absorbed by ice = m0*L0

(a) mass of ice melted (m0) = 1936.67/334 = 5.7984 g

(b) Volume of ice melted (V0) = 5.7984/0.917 = 6.3232 ml

(c) Volume of water that formed = 5.7984 ml

(d) Distance of bottom of penny from top ice surface = volume of ice melted/area of base = 6.3232/3.14*(0.9525)^2 = 2.2185 cm

So the distance from the center of the penny to the top ice surface = 2.2185 - 0.155/2 = 2.1410 cm

The water would not spill out as the volume that gets created when ice melts to water (6.3232-5.7984 = 0.5284 ml) is greater than the volume of the penny (0.4417 ml)

Hence the volume left after the penny sinks = 0.5284 - 0.4417 = 0.0831 ml = 0.0831 cm^3

The distance of the water level below the top ice surface = 0.0831/3.14*(0.9525)^2 = 0.0291 cm

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.