Solving for Ice: Coffee Cooling With Heat Exchange

AI Thread Summary
To solve for the amount of ice needed to cool 2.5 x 10^2 g of coffee from 70.0°C to 60.0°C, the heat exchange equations must be set equal, as the energy lost by the coffee equals the energy gained by the ice. The relevant equations include the heat gained by the ice, which involves both the latent heat of fusion and the specific heat capacity of water. The challenge lies in the two unknowns: the mass of ice and the total heat transfer (Q). It is crucial to account for the energy required to melt the ice and then heat the resulting water to the final temperature. Properly applying these principles will yield the correct mass of ice needed for cooling.
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Homework Statement


You are given 2.5 102 g of coffee (same specific heat as water) at 70.0°C. In order to cool this to 60.0°C, how much ice (at 0.0°C) must be added? Neglect heat content of the cup and heat exchanges with the surroundings.


Homework Equations


Q=(m ice)(c ice)(delta T ice+ m ice)(Lf)
Q=(m coffee)(c coffee)(delta T coffee)


The Attempt at a Solution


If these are the right formulas I still have the problem that there are 2 variables, since I don't know Q or m ice.
 
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The Q's in both equations are the same, energy lost by coffee = energy gained by ice.
so you simply set the two equations equal to eahc other
 
do you mean (m ice)(c ice)(delta T ice+ m ice)(Lf)=(m coffee)(c coffee)(delta T coffee)?
 
Almost, remember ther is energy to melt the ice (which doesn't depend on temperature) and then the energy to heat the resulting water to the final temperature.
 
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