- #1
JoshMG
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The question is:
An insulated beaker with negligible mass contains liquid water with a mass of 0.225 kg and a temperature of 75.8 C.
How much ice at a temperature of -23.6 C must be dropped into the water so that the final temperature of the system will be 31.0 C?
Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*k, and the heat of fusion for water to be 334 KJ/kg.
mice=?
This is what I have so far:
mw= 0.225 kg
Tw= 75.8 C
Ti= -23.6 C
mi= ?
Tf= 31.0 C
Cw= 4190 J/kg(K)
Ci= 2100 J/kg(K)
Lf= 334 KJ/kg = 334000J/kg (I would have to convert this right?)
With the information given I come up with these:
Qice warms= mi ci (0- (-5))
Qice melts= mi Lf
Qmix of waters= mi cw (31-0)
Qwater cools= mw cw (31 - 75.8)
Qwater cools = Qice warms + Qice melts + Q mix of waters
so now we can solve the mass of ice. right? But when I calculate it the answer I is really weird.
An insulated beaker with negligible mass contains liquid water with a mass of 0.225 kg and a temperature of 75.8 C.
How much ice at a temperature of -23.6 C must be dropped into the water so that the final temperature of the system will be 31.0 C?
Take the specific heat of liquid water to be 4190 J/kg*K, the specific heat of ice to be 2100 J/kg*k, and the heat of fusion for water to be 334 KJ/kg.
mice=?
This is what I have so far:
mw= 0.225 kg
Tw= 75.8 C
Ti= -23.6 C
mi= ?
Tf= 31.0 C
Cw= 4190 J/kg(K)
Ci= 2100 J/kg(K)
Lf= 334 KJ/kg = 334000J/kg (I would have to convert this right?)
With the information given I come up with these:
Qice warms= mi ci (0- (-5))
Qice melts= mi Lf
Qmix of waters= mi cw (31-0)
Qwater cools= mw cw (31 - 75.8)
Qwater cools = Qice warms + Qice melts + Q mix of waters
so now we can solve the mass of ice. right? But when I calculate it the answer I is really weird.
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