- #1
KurtWagner
- 43
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I must be doing something wrong here but I cannot figure out what it is1. Homework Statement
An insulated beaker with negligible mass contains a mass of 0.350kg of water at a temperature of 70.1∘C.
How many kilograms of ice at a temperature of − 11.6∘C must be dropped in the water to make the final temperature of the system 25.0∘C?
Take the specific heat for water to be 4190J/(kg⋅K) , the specific heat for ice to be 2100J/(kg⋅K) , and the heat of fusion for water to be 334kJ/kg .
Q = mcT
Q = mcT + Lfm
So basically I was just doing...
mwcw ΔTw = mi [(ci ΔTi) + Lf i]
which gives
ΔTw = 70.1∘C. - 25∘C = 45.1K
ΔTi = − 11.6∘C - 25∘C = 36.6K
0.350kg * 4190J/(kg⋅K) * (45.1K) = mi [(2100J/(kg⋅K)*(36.6K)] +334kJ/kg]
66139.15J = 257140J/kg * mi
mi = 0.161kg
which does not turn out to be the right answer (I do not have access to the right answer).
Can anyone tell me what mistake I am making.
Any help would be greatly appreciated.
An insulated beaker with negligible mass contains a mass of 0.350kg of water at a temperature of 70.1∘C.
How many kilograms of ice at a temperature of − 11.6∘C must be dropped in the water to make the final temperature of the system 25.0∘C?
Take the specific heat for water to be 4190J/(kg⋅K) , the specific heat for ice to be 2100J/(kg⋅K) , and the heat of fusion for water to be 334kJ/kg .
Homework Equations
Q = mcT
Q = mcT + Lfm
The Attempt at a Solution
So basically I was just doing...
mwcw ΔTw = mi [(ci ΔTi) + Lf i]
which gives
ΔTw = 70.1∘C. - 25∘C = 45.1K
ΔTi = − 11.6∘C - 25∘C = 36.6K
0.350kg * 4190J/(kg⋅K) * (45.1K) = mi [(2100J/(kg⋅K)*(36.6K)] +334kJ/kg]
66139.15J = 257140J/kg * mi
mi = 0.161kg
which does not turn out to be the right answer (I do not have access to the right answer).
Can anyone tell me what mistake I am making.
Any help would be greatly appreciated.