Physics final exam question: warm aluminum, cold ice heat transfer

Physics 205A Final Exam, Fall Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 14.39

[Version 1]
[10 points.] An ice sample (0.0° C) is placed onto a 5.00 kg aluminum block at 1.0° C. As a result, the entire ice sample melts to 0.0° C water. As this system reached a thermal equilibrium of 0.0° C, was heat transferred from the ice sample to the aluminum block, or from the aluminum block to the ice sample, or were both transfers simultaneously taking place? Explain your answer using the properties of heat, temperature, and thermal equilibrium.

Solution and grading rubric:

• p = 10/10:
  Correct. Heat transfers from warmer to cooler temperature objects. The aluminum temperature decreases, and thus is losing heat, while the ice melts (while its temperature remains constant), and thus is gaining heat. Discusses how each object loses or gains heat, such that the direction of the transfer is certain.
  
• r = 8/10:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May confuse heat with internal energy (heat transfers due to a difference in energy rather than temperature).
  
• t = 6/10:
  Nearly correct, but argument has conceptual errors, or is incomplete. Has heat flowing from aluminum to ice, but claims that internal energy of ice is unchanged (or Q_ice = 0) as its temperature remains constant.
  
• v = 4/10:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Has heat transferring from the ice to the aluminum, or has heat transferring from the aluminum causing the ice temperature to rise, or states that heat was transferred simultaneously from aluminum to ice and vice versa. At least has some discussion of temperature gradient, temperature/phase changes, and/or internal energy changes.
  
• x = 2/10:
  Implementation/application of ideas, but credit given for effort rather than merit. Statement of heat direction with little or substantive discussion.
  
• y = 1/10:
  Irrelevant discussion/effectively blank.
  
• z = 0/10:
  Blank.
  

Grading distribution:
Section 72177
p: 8 students
r: 0 students
t: 2 students
v: 1 student
x: 1 student
y: 1 student
z: 0 students

[Version 2]
[10 points.] An aluminum cylinder at room temperature (25.0° C) is dropped into a hole in a block of ice at 0.0° C. As this system reached a thermal equilibrium of 0.0° C, was heat transferred from the ice block to the aluminum cylinder, or from the aluminum cylinder to the ice block, or were both transfers simultaneously taking place? Explain your answer using the properties of heat, temperature, and thermal equilibrium.

Grading distribution:
Sections 70854, 70855
p: 28 students
r: 4 students
t: 8 students
v: 7 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 3334):

Another sample "p" response (from student 9626):

A sample "r" response (from student 0570):

A sample "t" response (from student 2468):

Another sample "t" response (from student 5256):

A sample "v" response (from student 1889):

Another sample "v" response (from student 4590):

Physics final exam problem: hot bullet into ice block

Physics 205A Final Exam, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 14.102

[20 points.] A 0.44 kg lead slug leaves a rifle at a temperature of 155 degrees C and travels at a speed of 190 m/s until it hits a 20.0 kg block of ice at 0 degrees C and comes to a rest within it. Determine how much ice will melt. Show your work and explain your reasoning.

(Specific heat of lead is 0.13 kJ/(kg*K); specific heat of ice is 2.1 kJ/(kg*K); latent heat of fusion for water is 333.7 (kJ/kg) ; specific heat of water is 4.19 kJ/(kg*K).)

Solution and grading rubric:

• p = 20/20:
  Correct. The kinetic energy of the bullet that is lost (7,942 J) as it comes to a complete stop, along with the heat it gives up as it cools down to 0 degrees C (8,866) is equal to the heat absorbed by the ice as it melts (16,808 J), and the resulting mass of ice melting is found to be 0.0504 kg.
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Does not include the kinetic energy of the bullet.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Sections 30880, 30881
p: 2 students
r: 5 students
t: 22 students
v: 9 students
x: 0 students
y: 0 students
z: 1 student

A sample "p" response (from student 5225):

A sample "r" response (from student 0008), with a kJ to J conversion error:

Another sample "r" response (from student 1830), with similar kJ to J conversion errors, but no explicit calculation for the amount of ice melted:

Another sample "r" response (from student 1990), with similar kJ to J conversion errors, coming to the conclusion that the entire ice block would melt:

Another sample "r" response (from student 6447), again with similar kJ to J conversion errors and concluding that the entire ice block would melt:

A sample "t" response (from student 1807), with no kinetic energy of the bullet:

Another sample "t" response (from student 1991), again with no kinetic energy of the bullet:

Physics final exam question: hot coffee plus frozen coffee

Physics 5A Final Exam, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 14.40

[20 points.] A Physics 5A student is going to make iced coffee by first brewing hot coffee, at a temperature of 85.0° C, and pouring it into a glass containing 0.300 kg of frozen coffee from a freezer at –15.0° C. How much hot coffee should the student pour into the glass, in order to result in a final temperature of 10.0° C? Assume that coffee is essentially water. Neglect the temperature change of the glass. Show your work and explain your reasoning.

(The specific heat of ice is 2.10 kJ/(kg*K); the specific heat of water is 4.19 kJ/(kg*K); the latent heat of fusion for water is 334 kJ/kg.)

Solution and grading rubric:

• p = 20/20:
  Correct. Net heat exchange with the environment (or glass) is set to zero, and equates this to sum of the heat taken in by the ice to warm up from -15.0 degrees C to 0 degrees C, the heat taken in to melt this ice completely, the heat taken in by the melted ice (as water) to warm up from 0 degrees C to 10 degrees C, and the heat given up by the coffee to cool down from 85.0 degrees C to 10.0 degrees C. Solves for the mass of the hot coffee, which is 0.389 kg.
  
• r = 16/20:
  Nearly correct, but includes minor math errors. May neglect the specific heat capacity for ice and water being different, or may have converted temperature changes in Celsius into absolute temperatures in Kelvins.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has energy balance equation with systematic application of Q = m*c*delta(T) and Q = m*L heat exchanges.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
p: 3 students
r: 3 students
t: 10 students
v: 15 students
x: 0 students
y: 1 student
z: 0 students

A sample of a "p" response (from student 1484) is shown below:
An "r" response (from student 7937) that uses the same heat capacity for ice and for liquid coffee:
An "x" response (from student 1337) that ends on a friendly note:

Physics clicker question: latent heat

Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 14.4

Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:

[0.6 participation points.] The heat capacity of a material undergoing a phase change (such as melting/freezing, or boiling/condensing) is typically:
(A) 0.
(B) ∞.
(C) some finite number.
(D) (Not enough information is given.)
(E) (I'm lost, and don't know how to answer this.)

Sections 4987, 4988
(A) : 8 students
(B) : 7 students
(C) : 12 students
(D) : 1 student
(E) : 0 students

Correct answer: (B)

The heat capacity is given by:

C = Q/delta(T).

When heat is put into (or taken from) a system undergoing a phase change, the temperature remains constant, and thus the denominator is zero, making the heat capacity infinite. Thus a system undergoing a phase change cannot have a defined heat capacity, as the heat goes into breaking or making bonds, instead of increasing/decreasing the amount of thermal energy in the system. Instead, the latent heat is defined:

L = Q/m,

where m is the amount of material undergoing a phase change.