# Fine temperature of equal masses of hot tea and ice after melting

• Jayhawk1
In summary, the final temperature of iced tea is dependent on the initial hot tea temperature. If it is at 94C, the final temperature will be 14.45C. If the initial hot tea is at 65C, the final temperature will be 5.59C.
Jayhawk1
A person makes a quantity of iced tea by mixing 559 g of hot tea (essentially water) with an equal mass of ice at its melting point. If the initial hot tea is at a temperature of 94 C, what are the (a) final temperature of the ice tea (in oC) and (b) mass of the remaining ice? what are the (c) final temperature of the ice tea (in oC) and (d) mass of the remaining ice if the initial hot tea is at a temperature of 65 C? The specific heat of water is 4186 J/(kg.oC) and the latent heat of fusion is 333 kJ/kg.

Now... all I can get is B--- I know that's zero since the ice will all melt. How do I figure out the rest?? Any help would be appreciated?

Jayhawk1 said:
A person makes a quantity of iced tea by mixing 559 g of hot tea (essentially water) with an equal mass of ice at its melting point. If the initial hot tea is at a temperature of 94 C, what are the (a) final temperature of the ice tea (in oC) and (b) mass of the remaining ice? what are the (c) final temperature of the ice tea (in oC) and (d) mass of the remaining ice if the initial hot tea is at a temperature of 65 C? The specific heat of water is 4186 J/(kg.oC) and the latent heat of fusion is 333 kJ/kg.

Now... all I can get is B--- I know that's zero since the ice will all melt. How do I figure out the rest?? Any help would be appreciated?

Latent heat of fusion represents the amount of heat ice must absorb to change into liquid without changing temperture. Once it becomes water, it will warm up at the same rate (amount of heat per temperature change) as an equal amount of water will cool down. Since you have equal amounts of ice and water in this problem, once the ice is melted the mixture will warm to a temperature half way between 0C and the temperature the water will have at the moment the ice is all melted. Figure out how much the water temperature has to drop to just melt the ice. Then find the final temperature as the ice water warms and the warmer water cools.

For C and D, see if there is enough heat released by the cooling water to melt all the ice. If not, how much ice can be melted? If all of it, the problem becomes just like A and B. If not, the final temperature is ice temperature with some ice remaining. In either case you have

Latent heat + warming heat (if any) = cooling heat

I'm sorry, but my professor doesn't really teach very well and I don't even know where to start even with your very thorough help... can you be a bit more specific? Thanks.

Suppose you have 1kg of ice. It would take 333 kJ to melt it leaving it at a temperature of 0C. 1kg of water will lose or gain 4.186 kJ of heat for every temperature change of 1C. To give up 333 kJ of heat, 1kg of water would have to cool 79.55C. In the first part of the problem, water can cool 94C, so it is hot enough to melt all the ice and still have a temperature of 94-79.55=14.45C. The melted ice at 0C and this cooled water at 14.45C will mix with the warmer water giving heat to the cooler water until the temperature is the same. Since there are equal amounts at each temperature, the final temperature will be half way between.

In the second part, the hot water is not hot enough to melt all the ice. The water can only cool 65C. You need to figure out how much heat the water will lose dropping from 65C to 0C using the 4.186 kJ/(kg.oC). Then using the 333 kJ/kg latent heat to melt ice, figure out how much ice will melt if that much heat is absorbed. The final temperature will be 0C with a lot, but not all of the ice melted.

In general, the masses need not be that same. What you have to look at is heat lost = heat gained. If m is the mass if ice and M is the mass of water then

m*333kJ/kg + m*(Temp rise)*4.186 kJ/(kg.oC) = M*(Temp drop)*4.186 kJ/(kg.oC)

with the temp rise and temp drop leading to the same temperature throughout. The temperature cannot drop below zero, so in some cases there is not temperature rise and the final temperature is zero.

So what do I put in for temp rise and drop, 65?

Jayhawk1 said:
So what do I put in for temp rise and drop, 65?

It turns out to be 65C for the second part because it can't go below zero,

For the first part

Temp rise = Temp final - 0C
Temp drop = 94C - Temp final

You need to solve for Temp final

For the second part... why won't it work. Temp rise is 0, and Temp drop is 65, correct? If so... my answer isn't right.

Got it! Thanks so much!

Last edited:
notorious big said:
For the second part... why won't it work. Temp rise is 0, and Temp drop is 65, correct? If so... my answer isn't right.

Probably because you are using the total mass of the ice on the left side. When you know the final temperature is zero because not all of the ice melts, the m on the left side is the mass of the ice that does melt, and that is what you are trying to find.

## 1. What is the fine temperature of equal masses of hot tea and ice after melting?

The fine temperature of equal masses of hot tea and ice after melting is approximately 0 degrees Celsius or 32 degrees Fahrenheit. This is the point at which the ice and hot tea reach a thermal equilibrium, meaning they are both at the same temperature.

## 2. Why does the temperature of hot tea decrease when ice is added?

This is because the ice absorbs heat from the hot tea in order to melt. As the ice melts, it takes in heat energy from the hot tea, causing the overall temperature to decrease.

## 3. How does the rate of melting ice affect the temperature of the hot tea?

The rate of melting ice does not directly affect the temperature of the hot tea. The temperature will continue to decrease until the ice has completely melted and the two substances reach thermal equilibrium.

## 4. What factors can affect the fine temperature of equal masses of hot tea and ice after melting?

The main factors that can affect the fine temperature of equal masses of hot tea and ice after melting are the starting temperature of the hot tea, the amount of ice added, and the thermal properties of the container holding the substances.

## 5. Is the fine temperature of equal masses of hot tea and ice after melting always 0 degrees Celsius or 32 degrees Fahrenheit?

No, the fine temperature can vary depending on the specific conditions. For example, if the starting temperature of the hot tea is lower, it may take longer for the two substances to reach thermal equilibrium and the final temperature may be slightly higher. Additionally, the type and amount of ice added can also impact the final temperature.

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