Physics quiz question: sub-zero ice versus super-zero water

Physics 8A Quiz 11, Spring Semester 2007
Cuesta College, San Luis Obispo, CA

Physics 8A learning goal Q11.2

[3.0 points.] Consider an ice cube of unknown mass that is taken from a freezer, where the cube's temperature was –5.00 degrees C, and dropped into an insulated container of water (mass unknown) at +5.00 degrees C. Assume that no heat was transferred to/from the contents of the container while its contents reach an equilibrium state. c_ice = 2,100 J/(kg*K). c_water = 4,186 J/(kg*K). L_fusion/melting = 334,000 J/kg.

(Cf. Young and Freeman, University Physics, 11/e, Problem 17.101(a).)

[Version 1]

Which one of the following choices best describes the initial mass of –5.00 degrees C ice, compared to the initial mass of +5.00 degrees C water, if the final equilibrium state is entirely water at 0 degrees C?
(A) m_ice < m_water.
(B) m_ice = m_water.
(C) m_ice > m_water.
(D) (Not enough information is given to determine the relative mass of ice and water in the initial state of the system.)

Correct answer: (A)
To warm up from -5.00 degrees C to 0 degrees C, 10,500 J of heat must be added per kg of ice. To cool down from +5.00 degrees C to 0 degrees C, 20,930 J of heat must be removed per kg of water. Thus m_ice = m_water, then the final state would be a mixture of ice and water at 0 degrees C. Since the final state is all water at 0 degrees C, then 334,000 J of heat must have been added per kg of ice to melt it, and thus m_ice < m_water.

Student responses:
(A) : 3 students
(B) : 3 students
(C) : 9 students
(D) : 0 students

[Version 2]

Which one of the following choices best describes the initial mass of –5.00 degrees C ice, compared to the initial mass of +5.00 degrees C water, if the final equilibrium state is entirely ice at 0 degrees C?
(A) m_ice < m_water.
(B) m_ice = m_water.
(C) m_ice > m_water.
(D) (Not enough information is given to determine the relative mass of ice and water in the initial state of the system.)

Correct answer: (C)
To warm up from -5.00 degrees C to 0 degrees C, 10,500 J of heat must be added per kg of ice. To cool down from +5.00 degrees C to 0 degrees C, 20,930 J of heat must be removed per kg of water. Thus m_ice = m_water, then the final state would be a mixture of ice and water at 0 degrees C. Since the final state is all ice at 0 degrees C, then 334,000 J of heat must have been removed per kg of water to freeze it, and thus m_ice > m_water.

Student responses:
(A) : 2 students
(B) : 1 student
(C) : 10 students
(D) : 0 students

Another one of those questions with two different versions with non-complementary distribution of responses...