Ice cube placed in water - find final temperature

• knownothing1
In summary, the conversation is discussing a problem where a 75 gram ice cube at 0 degrees Celsius is placed in 825 grams of water at 25 degrees Celsius and the final temperature of the mixture is being asked. The formula being used is (75g)(2.108kJ/kg-k)(T2-0)=(825g)(4.184J/g*degrees C)(T2-25 degrees C), but there may be issues with mixed units and not accounting for latent heat from melting the ice. The expected answer is 16 degrees Celsius.
knownothing1

Homework Statement

A 75 gram ice cube at 0 degrees Celcius is placed in 825 grams of water at 25 degrees C. What is the final temperature of the mixture?

Homework Equations

What is the best formula to use in this question?

The Attempt at a Solution

I have attempted this problem multiple times. The last time I plugged in these variables:
(75g)(2.108kJ/kg-k)(T2-0)=(825g)(4.184J/g*degrees C)(T2-25 degrees C)
The answer is supposed to be 16 degrees C. I can't seem to figure this one out. Any help will be greatly appreciated. Am I missing variables?

Well, I can think of two things you might check on the left side of your equation you seem to have mixed units you have 75g of ice but write J/kg.

Also, don't you have to melt the ice and wouldn't that involve latent heat?

I would approach this problem by first identifying the relevant equations and variables. The equation that relates heat, mass, and temperature change is q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we have two substances - the ice cube and the water - with different masses and initial temperatures, and we want to find the final temperature of the mixture.

To solve this problem, we can first calculate the heat lost by the ice cube as it warms up from 0 degrees Celcius to its final temperature, using the equation q = mcΔT. Since the ice cube is initially at 0 degrees Celcius, its specific heat capacity is 2.108 kJ/kg-K. Plugging in the values, we get q = (75g)(2.108kJ/kg-K)(Tf-0), where Tf is the final temperature of the mixture.

Next, we can calculate the heat gained by the water as it cools down from 25 degrees Celcius to the same final temperature, using the same equation q = mcΔT. Since the water is initially at 25 degrees Celcius, its specific heat capacity is 4.184 J/g-°C. Plugging in the values, we get q = (825g)(4.184J/g-°C)(Tf-25).

Since the heat lost by the ice cube is equal to the heat gained by the water (assuming no heat is lost to the surroundings), we can set the two equations equal to each other and solve for Tf. This will give us the final temperature of the mixture. So, our final equation is (75g)(2.108kJ/kg-K)(Tf-0) = (825g)(4.184J/g-°C)(Tf-25).

Solving for Tf, we get Tf = 16 degrees Celcius, as expected. Therefore, the final temperature of the mixture is 16 degrees Celcius. It is important to note that when solving problems like this, it is crucial to pay attention to units and conversions, as they can greatly affect the final answer.

1. How does the temperature of the water change when an ice cube is placed in it?

When an ice cube is placed in water, the temperature of the water will decrease as the ice cube absorbs heat from the water and melts.

2. What factors affect the final temperature when an ice cube is placed in water?

The final temperature of the water will be affected by the temperature of the water before the ice cube is added, the mass and temperature of the ice cube, and the amount of water present.

3. How can you calculate the final temperature when an ice cube is placed in water?

The final temperature can be calculated using the equation Qw = -Qice, where Qw is the heat gained by the water and Qice is the heat lost by the ice. This equation can be rearranged to solve for the final temperature, which is the point where the two heat values are equal.

4. Why does the final temperature of the water and ice cube system not equal the average of the initial temperatures?

The final temperature is not the average of the initial temperatures because the heat transfer between the ice cube and water is not instantaneous. The ice cube will continue to melt and absorb heat until it reaches the same temperature as the water, resulting in a lower final temperature than the average of the initial temperatures.

5. How does the volume of the ice cube affect the final temperature of the water?

The volume of the ice cube affects the final temperature because a larger ice cube will take longer to melt and absorb heat, resulting in a lower final temperature compared to a smaller ice cube. This is because a larger ice cube has a larger surface area and thus a larger area for heat transfer to occur.

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