Ice cube placed in water - find final temperature

[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?

 

[AEM]

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?

 

[kerimek]

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.