Heat transfer calculation problem

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Homework Help Overview

The problem involves calculating the mass of ice required to achieve a specific final temperature in a system containing water. The context is heat transfer, specifically focusing on the thermal interactions between water and ice at different temperatures.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the heat transfer equations and the assumptions made regarding the state changes of ice and water. Questions are raised about the correct application of the heat transfer equations and the behavior of ice as it warms and melts.

Discussion Status

Participants are actively exploring the problem, with some providing insights into the necessary steps for the heat transfer calculations. There is recognition of the need to account for the phase change of ice and its specific heat after melting, indicating a productive direction in the discussion.

Contextual Notes

There is an acknowledgment of potential misunderstandings regarding the heat transfer process, particularly the transitions of ice to water and the specific heat values involved. The original poster expresses uncertainty about their calculations and seeks clarification on their approach.

KurtWagner
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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 .

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.
 
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What happens to the ice when it reaches a temperature of 0 C? Is your equation equating the heat lost by the water in the beaker to the heat gained by the ice dropped into the beaker correct? You should first write out what happens to the ice after it is dropped into the beaker without applying any numbers.
 
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Q = mi [(ci ΔTi) + Lf i]

Would this equation not mean that each kilo of ice absorbs the heat needed to change its state and also to change its temp?
for each kilo 2100 Joules per degree change in temp but also the 334kJ to change its state?

Am i right in assuming that each kilo of ice changes through -11.6∘C to 0∘C, taking 24360J , changing state for 334000J, then continuing to change from 0∘C to 25∘C taking a further 52500J. Total = 410860J/kg

Whoops, that's different from above, but It is what I used in my calculations
 
oh... It becomes water and has a different specific heat after 0∘C... hmmmm. Ill try that
 
That worked.

Qwater = Qice + Qicewater
 

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