Solution to Thermo Ice Melting Problem

AI Thread Summary
To determine how much ice at -15.0 C must be added to 0.325 kg of water at 69.9 C to achieve a final temperature of 39.0 C, the heat transfer equations are applied. The calculations involve the heat required to melt the ice, heat the melted ice to 39.0 C, and the cooling of the water from 69.9 C to 39.0 C. The initial solution yields a mass of ice of 0.2155 kg, but there is confusion regarding the energy needed for melting the ice, which is 334 kJ/kg. The correct approach emphasizes including the heat of fusion in the calculations to ensure accurate results.
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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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