Solution to Thermo Ice Melting Problem

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SUMMARY

The discussion centers on solving the Thermo Ice Melting Problem involving an insulated beaker containing 0.325 kg of water at 69.9°C and determining the mass of ice at -15.0°C needed to achieve a final temperature of 39.0°C. The specific heats of water and ice are given as 4190 J/(kg·°C) and 2100 J/(kg·°C), respectively, with the heat of fusion for ice being 334,000 J/kg. The calculated mass of ice required is 0.2155 kg, but participants noted the importance of including the energy needed to melt the ice in their calculations.

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Homework Statement


An insulated beaker with negligible mass contains liquid water with a mass of 0.325 \rm{kg} and a temperature of 69.9 C.

How much ice at a temperature of -15.0 C must be dropped into the water so that the final temperature of the system will be 39.0 C?
Take the specific heat of liquid water to be 4190 , the specific heat of ice to be 2100 , and the heat of fusion for water to be 334

Homework Equations


Q = mcT
Q = mL

The Attempt at a Solution


m = mass of ice

Melt Ice = 334 * m
Heat Ice = 2100 * (0--15) * m
Heat ice water = m*4190 * (39-0)

Cool Water = 0.325 * 4190 (39-69.9) = 42078.075

Solve for m
42078.075 = 334 * m + 2100 * (0--15) * m + m*4190 * (39-0)

m= 0.2155 kgFound another equation and used it and got the same answer... not sure where I am going wrong.

Mice = Mwater * Cwater (Tf - Twater)/(Cice * Tice -Lf - Cwater * Tf)

[0.325 * 4190 * (39-69.9)] / (2100 * -15 -334-4190*39) =0.2155kg
 
Last edited:
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Anyone got an idea what's up? When I plugged in the answer it told me "Dont forget the energy needed to melt the ice. ?
 
The heat of fusion if Ice is actually 334 KJ/kg or 334*10^3 J/kg
 

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