There are 1050 grams of water in a container that has insulating sides and a hea
ID: 881301 • Letter: T
Question
There are 1050 grams of water in a container that has insulating sides and a heat-conducting base. The container is placed on a variable hot plate, the temperature of which is adjusted so that the heat is added to the water at a constant rate of 760 W. The base of the container has an area of 100 cm^2. The material of the base has thermal conductivity of 353 W/m*K.
a. How long does it take for the temperature of this water to go from 20 degrees C to the boiling point?
b. How long after the water first starts to boil will one third of it be boiled away?
While the water is boiling, the hot plate temperature must be 106 degrees C to maintain the 760 W heat flow.
c. How thick is the base of the container?
220 grams of the remainining boiling water is now poured into another insulated container holding water at 20 degrees C. The final temperature of this mixture is 63.2 degrees C.
d. What mass of water was in this container?
The rest of the boiling water is poured into an insulated jar holding ice at 0 degrees C. The final temperature of this mixture is 51.0 degrees C.
e. What mass of ice was in the jar?
Explanation / Answer
m = 1050g
Q = 760 W
A = 100 cm2
k = 353 W/m*K
Questions:
a. How long does it take for the temperature of this water to go from 20 degrees C to the boiling point?
Q = mCp(Tf-Ti)
Boiling point Tf = Tb = 100C and Cp of Water = 4.18 J/gC
Q = 1050*4.18*(100-20) = 351120 J are required
Since Power = Q/t = 760 J/s
t = Q/P = 351120/760 = 462 seconds
t = 7.7 minutes
b. How long after the water first starts to boil will one third of it be boiled away?
For 1/3 of boiled material:
Q = latent heat
Q = m*lh
But we want 1/3 of mass so:
Q = 1/3*m*lh
lh = 2260 J/g
Q = 1/3*1050*2260 = 791000 J are required
since P = Q/t
then t = Q/P = 791000/760 = 1040.8 sec or 17.35 minutes
t = 17.35 minutes
While the water is boiling, the hot plate temperature must be 106 degrees C to maintain the 760 W heat flow.
c. How thick is the base of the container?
For material thickness:
q = k A dT / s
q = Watts
k = conductivity
A = m2
dT = (T2-T1)
s = material thickness
Solve for s
q = k A dT / s
s = k A dT / q
s = 353 W/m*K * (100 cm2)(1m2/100^2 cm2)*(106-100)C /(760W)
s = 353*0.01*(6)/(760) = 0.02786 m
s = 2.78 cm
220 grams of the remainining boiling water is now poured into another insulated container holding water at 20 degrees C. The final temperature of this mixture is 63.2 degrees C.
d. What mass of water was in this container?
-Q lost = Qwin
Q lost = m*Cp*(Tf-Ti) --- NOTE: assume its water at 100ºC
Q lost = 220*4.18*(63.2-1000) = -33841.3 J
Q win = m*Cp(Tf-Ti) = m*4.18*(63.2-20)
Since Q win = -Qlost
33841.3 = m*4.18*(63.2-20)
m = 33841.3/(4.18*(63.2-20)) = 187.4 g of "Cold" Water
m = 187.4 g of "Cold" Water
The rest of the boiling water is poured into an insulated jar holding ice at 0 degrees C. The final temperature of this mixture is 51.0 degrees C.
e. What mass of ice was in the jar?
Boiling Water = 0 g since there were all used
BUT assume 1050 - 220 = 830 g of boiling water where used
Qlost = mCp(Tf-Ti)
Qlost = 830*4.18*(51-100) = -170,000.6 J
Qwin = Qmelt + Qwater
Qwin = -Qlost = 170,000.6
NOTE: HL of Ice = 2336
Qmelt = m*hl = m*336 = 336*m
Qwater = m*Cp(Tf-Ti) = m*4.18*(51-0) = 213.2*m
Qwin = Qmelt + Qwater = 336*m+213.2*m = 549.2*m
170,000.6 = 549.2*m
m = 309 grams of Ice
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