Equilibrium Temperature of Cup and Water When Adding Ice - Homework Solution

In summary, the problem involves finding the equilibrium temperature of a 35 g ice cube added to 110 g of water in a 62 g iron cup. The cup and water have an initial temperature of 36°C. The equations used for this problem are heat lost by water+cup, amount of heat needed to melt ice, amount of heat left, and deltaT. After plugging in the numbers, a miscalculation error was found in converting units. The correct solution was obtained after correcting this error.
  • #1
map7s
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Homework Statement


A 35 g ice cube at 0.0°C is added to 110 g of water in a 62 g iron cup. The cup and the water have an initial temperature of 36°C. Find the equilibrium temperature of the cup and its contents.


Homework Equations


heat lost by water+cup Q(w)=m(water)c(w)deltaT+m(cup)c(c)deltaT
amt. of heat needed to melt ice Q(ice)=m(ice)L(f)
amt. of heat left Q=Q(w)-Q(ice)
deltaT=Q/((m(w)+m(ice))c(w)+(m(cup))c(cup))

The Attempt at a Solution


I tried plugging in my numbers to the equations above but I didn't get the right answer. I'm fairly certain that my equations are correct but the only one that I am uncertain about is the last one.
 
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  • #2
Never mind...I found out that it was simply a miscalculation error in converting units. :)
 
  • #3
I'm not sure if the deltaT should be positive or negative in this case.

I would like to first clarify that the provided equations are not entirely accurate and may lead to incorrect results. The correct equations for this problem would be:

Heat lost by water and cup: Q = (mwater * cwater + mcup * ccup) * ΔT

Heat needed to melt ice: Q = mice * Lf

Equilibrium temperature: Q = 0, which means that the heat lost by the water and cup must be equal to the heat needed to melt the ice.

Now, plugging in the given values, we get:

Heat lost by water and cup: Q = (110 g * 4.18 J/g°C + 62 g * 0.45 J/g°C) * (36°C - Tf)

Heat needed to melt ice: Q = 35 g * 334 J/g

Equilibrium temperature: (110 g * 4.18 J/g°C + 62 g * 0.45 J/g°C) * (36°C - Tf) = 35 g * 334 J/g

Solving for Tf, we get Tf = -0.46°C. This means that the equilibrium temperature of the cup and its contents will be slightly below 0°C, which is the melting point of ice. This makes sense because the ice will absorb heat from the water and cup, causing them to cool down and reach an equilibrium temperature below the initial temperature of 36°C.

Therefore, it is important to use the correct equations and units when solving scientific problems to ensure accurate results. Additionally, it is always a good practice to double check your calculations and results to avoid any errors.
 

1. Why does ice float in water?

Ice floats in water because it is less dense than liquid water. When water freezes, its molecules form a crystalline structure that takes up more space than the same amount of liquid water. This results in ice having a lower density and causing it to float on top of the denser liquid water.

2. How does salt affect the melting point of ice in water?

Salt lowers the melting point of ice in water. When salt is added to water, it disrupts the hydrogen bonding between water molecules, making it harder for ice to form. This causes the freezing point of the water to decrease, making it more difficult for the ice to stay solid and causing it to melt at a lower temperature.

3. What other factors can affect the rate of ice melting in water?

Other factors that can affect the rate of ice melting in water include temperature, air pressure, and the concentration of impurities in the water. Higher temperatures and lower air pressure can both cause ice to melt faster, while impurities in the water can lower the melting point and cause it to melt more quickly.

4. Does the shape or size of ice affect how quickly it melts in water?

Yes, the shape and size of ice can affect how quickly it melts in water. Smaller pieces of ice will have a larger surface area in contact with the water, causing them to melt faster. Similarly, ice with a larger surface area, such as crushed ice, will melt more quickly than a larger solid block of ice.

5. How does the presence of other substances, like oil, affect the melting of ice in water?

The presence of other substances, such as oil, can affect the melting of ice in water. Oil is less dense than water, so when it is present on the surface of the water, it can form a barrier that prevents heat from reaching the ice. This can slow down the melting process, causing the ice to take longer to melt in oil-contaminated water compared to pure water.

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