Mass of Ice required to achieve a certain final Temperature

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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 scenario includes an insulated beaker with water at a higher temperature and ice at a lower temperature, with the goal of reaching thermal equilibrium at 40.0 °C.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the energy balance equation involving heat transfer between the ice and water. There is an attempt to identify potential errors in the calculations and notation used in the problem.

Discussion Status

The discussion is ongoing, with participants exploring the implications of variable notation and its impact on the calculations. Some guidance has been offered regarding the importance of clear variable definitions, and there is recognition of a previously overlooked piece of information.

Contextual Notes

Participants note that the problem involves specific heat capacities and the heat of fusion, and there is an emphasis on careful notation to avoid confusion in calculations.

CrosisBH
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Homework Statement


An insulated beaker with negligible mass contains a mass of 0.350 kg of water at a temperature of 76.5 °C.
How many kilograms of ice at a temperature of − 23.9 ∘C must be dropped in the water to make the final temperature of the system 40.0 ∘C?
Take the specific heat for water to be 4190 J/(kg⋅K) , the specific heat for ice to be 2100 J/(kg⋅K) , and the heat of fusion for water to be 334 kJ/kg .

Homework Equations



Q = mc\Delta T
Q = mL_f

The Attempt at a Solution


I recognized what exactly happened which each part of the system.

Ice at T_ice is heated to 0°C -> Ice is transformed to water -> The water is heated to T_f
Water at T_water is cooled to T_f

So setting up the equation
Q_{ice} + Q_{ice\rightarrow water} + Q_{water(ice)} + Q_{water} = 0
Expanding
m_{ice} c_{ice} (0°C - T_{ice}) + m_{ice}L_f + m_{ice}c_{water}(T_f - 0°C)+m_{water}c_{water}(T_f-T_{water}) = 0
I then solved for m_ice,
m_{ice} = \frac{m_{water} c_{water} T_i - m_{water} c_{water} T_f}{L_f + c_{water} T_f - c_{ice} T_{ice}}

And then I plugged all the values into Desmos:
upload_2019-1-25_21-8-23.png


L_f was converted from kJ/kg to J/kg. Getting -0.170 kg when doing 3 significant figures. Obviously there's something wrong here. I went through the algebra multiple times, so I think there's something wrong with my Physics.
 

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CrosisBH said:
something wrong with my Physics.
No, just lack of care in notation. Do you notice there's a given number you have not used?
 
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haruspex said:
No, just lack of care in notation. Do you notice there's a given number you have not used?
Yep. It was bad notation. Dumb me used T_i as both temperature of ice and the initial temperature of water in Desmos. I guess this is a lesson for me to properly name my variables so no confusion can stir from me lol.
 
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Likes   Reactions: Tom.G
Well I find out something new everyday. That Desmos looks impressive. It may be overkill for this type of math problem. Perhaps that is how one of your variables got lost.
 

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