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ID: 1408721 • Letter: H
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home / study / science / physics / questions and answers / (a) in air at 0°c, a 1.72-kg copper block at 0°c ...
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(b) As the block slows down, identify its energy input Q, its change in internal energy Eint, and the change in mechanical energy for the block–ice system.
(c) For the ice as a system, identify its energy input Q and its change in internal energy Eint.
(d) A 1.72-kg block of ice at 0°C is set sliding at 2.55 m/s over a sheet of copper at 0°C. Friction brings the block to rest. Find the mass of the ice that melts.
home / study / science / physics / questions and answers / (a) in air at 0°c, a 1.72-kg copper block at 0°c ...
Question
(b) As the block slows down, identify its energy input Q, its change in internal energy Eint, and the change in mechanical energy for the block–ice system.
(c) For the ice as a system, identify its energy input Q and its change in internal energy Eint.
(d) A 1.72-kg block of ice at 0°C is set sliding at 2.55 m/s over a sheet of copper at 0°C. Friction brings the block to rest. Find the mass of the ice that melts.
____mg
(e) Evaluate Q and Eint for the block of ice as a system and Emech for the block–ice system.
(f) Evaluate Q and Eint for the metal sheet as a system.
(g) A thin, 1.72-kg slab of copper at 21.0°C is set sliding at 2.55 m/s over an identical stationary slab at the same temperature. Friction quickly stops the motion. Assuming no energy is transferred to the environment by heat, find the change in temperature of both objects. (Assume the specific heat of copper is 387 J/kg · °C.)
(h) Evaluate Q and Eint for the sliding slab and Emech for the two-slab system.
(i) Evaluate Q and Eint for the stationary slab.
Q = ____J Eint = ____J Emech = ____JExplanation / Answer
The kinetic energy of copper block is given as
K.E = 0.5mV2
K.E = 0.5 (1.72)(2.55)2
K.E = 5.59J
As heat absorbed would be
Q = Mice * Lf Mice = mass of ice , Lf = Latent heat of fusion
Mice = Q / Lf
Mice = 5.59 / (3.3 * 105)
Mice = 0.000016Kg
Mice = 16.94mg
b) For the block
Q = 0 J
dEint= 0J
dEmech = -5.59J
c) For Ice
Q = 0
dEint = 5.59J
dEmec = 0J
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