A 5 kg block of ice at -20 degrees Celsius is added to an insulated contained pa

Question

A 5 kg block of ice at -20 degrees Celsius is added to an insulated contained partly filled with 10 kg of water at 20 degrees Celsius. A) Find the final temperature of the interior of the container. B) how much ice remains if any? How much more ice cold have been melted? Show work A 5 kg block of ice at -20 degrees Celsius is added to an insulated contained partly filled with 10 kg of water at 20 degrees Celsius. A) Find the final temperature of the interior of the container. B) how much ice remains if any? How much more ice cold have been melted? Show work

Explanation / Answer

first part is tricky because the ice does not
entirely melt in this example. When there is any doubt concerning
whether there will be a complete phase change,
some preliminary calculations are necessary. First, find
the total energy required to melt the ice, Q melt, and then
find Q water, the maximum energy that can be delivered by

the water above -20°C. If the energy delivered by the water is
high enough, all the ice melts. If not, there will usually be
a final mixture of ice and water at -20°C, unless the ice starts
at a temperature far below -20°C, in which case all the liquid
water freezes.

First, compute the amount of energy necessary to completely
melt the ice:
Qmelt = miceLf =5 (5.00 kg)(3.33 3 105 J/kg)
=1.67x106 J
Next, calculate the maximum energy that can be lost by
the initial mass of liquid water without freezing it:
Qwater=mwatercT
=(10.0 kg)(4 190 J/kg ? °C)(-20°C-20.0°C)

=-1.676x105J

This result is less than half the energy necessary to melt
all the ice, so the final state of the system is a mixture of
water and ice at the freezing point:
T =20°C
(b) Compute the mass of ice melted.
Set the total available energy equal to the heat of fusion
of m grams of ice, mLf , and solve for m:
1.676x105J = mLf = m(3.33 x105 J/kg)
1.676x105J =m(3.33 x105 J/kg)

m=0.503 kg

calculation may be wrong but concept is correct

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