My temperature after dropping ice in tea isn't adding up

[Eclair_de_XII]

Homework Statement

"If a cup of tea (175.0 mL, 90.0°C) is cooled by adding an ice cube (25.0 g, -15.0°C), what is the final temperature of the tea? The heat required to melt ice (heat of fusion) is 80.0 cal/g."

Homework Equations

S.H. of water (solid) = 0.5 cal/g°C

The Attempt at a Solution

1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = 0.5 cal/1 g°C ⋅ 25.0 g ⋅ 15.0°C + 80 cal/1 g ⋅ 25.0 g
175x - 15750 = 187.5 + 2000.0
175x = 17937.5
x = 102.5°C

This is not adding up. The final temperature should be lower than 90.0°C, not higher.

 

[collinsmark]

Homework Statement

"If a cup of tea (175.0 mL, 90.0°C) is cooled by adding an ice cube (25.0 g, -15.0°C), what is the final temperature of the tea? The heat required to melt ice (heat of fusion) is 80.0 cal/g."

Homework Equations

S.H. of water (solid) = 0.5 cal/g°C

The Attempt at a Solution

1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = 0.5 cal/1 g°C ⋅ 25.0 g ⋅ 15.0°C + 80 cal/1 g ⋅ 25.0 g
175x - 15750 = 187.5 + 2000.0
175x = 17937.5
x = 102.5°C

This is not adding up. The final temperature should be lower than 90.0°C, not higher.

[Changed some text color.]

Right. If x is less than 90.0^o it would make the term in red negative, meaning that the whole left hand side of the equation would be negative. Are you sure you want to set the problem up that way?

 

[NascentOxygen]

The water in the ice cube starts off at -15°, but what is its final temperature here?

 

[collinsmark]

Also, you need an additional term on the right hand side of the equation somewhere. Once the ice finishes melting, that corresponding water, that was once ice, must rise to temperature x. That needs to be represented in there somewhere. [Edit: NascentOxygen beat me to it. ]

 

[Chestermiller]

Yes, as Collinsmark is saying, the left hand side of the equation is supposed to be the amount of heat lost by the water to the ice, not the amount of heat gained by the water. Also, on the right hand side of the equation, you are missing a term representing the heat required to raise the temperature of the water from the melted ice to the final temperature.

EDIT: Oops. Double scooped.

Chet

 

[tech99]

Homework Statement

"If a cup of tea (175.0 mL, 90.0°C) is cooled by adding an ice cube (25.0 g, -15.0°C), what is the final temperature of the tea? The heat required to melt ice (heat of fusion) is 80.0 cal/g."

Homework Equations

S.H. of water (solid) = 0.5 cal/g°C

The Attempt at a Solution

1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = 0.5 cal/1 g°C ⋅ 25.0 g ⋅ 15.0°C + 80 cal/1 g ⋅ 25.0 g
175x - 15750 = 187.5 + 2000.0
175x = 17937.5
x = 102.5°C

This is not adding up. The final temperature should be lower than 90.0°C, not higher.

I think you need to equate the total heat at the start (tea-cube-latent heat) to that at the end when melting stops.
I think your equation is slightly adrift. The total heat at the start is that of the hot tea, minus that of the cube getting up to melting point, minus the latent heat. The heat at the end is the new volume of tea x unknown temp. The heat before and after is equal, so you can make an equation.

 

[Eclair_de_XII]

Are you sure you want to set the problem up that way?

I'm told by my textbook that I have to make everything on the right-hand side negative. In any case:
1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = -(0.5 cal/1 g°C ⋅ 25.0 g ⋅ (x + 15.0°C) + 80 cal/1 g ⋅ 25.0 g)
175x - 15750 = -12.5x - 187.5 - 2000
187.5x = 13562.5
x = 72.3333... °C

The water in the ice cube starts off at -15°, but what is its final temperature here?

So it should be: "1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = -(0.5 cal/1 g°C ⋅ 25.0 g ⋅ (x + 15.0°C) + 80 cal/1 g ⋅ 25.0 g)"?

Yes, as Collinsmark is saying, the left hand side of the equation is supposed to be the amount of heat lost by the water to the ice, not the amount of heat gained by the water.

I was actually sure that was what I was describing with the left-hand side of the equation.

 

[Eclair_de_XII]

The heat before and after is equal, so you can make an equation.

I'm honestly not sure what you mean.

 

[Chestermiller]

What's the heat capacity of the liquid water that forms after the ice melts? Is it the same as the heat capacity of the ice? If not, then please reconsider this term: 0.5 cal/1 g°C ⋅ 25.0 g ⋅ (x + 15.0°C)

Chet

 

[Eclair_de_XII]

So: 1 cal/1 g°C ⋅ 25.0 g ⋅ (x - 15.0°C)?

 

[Chestermiller]

So: 1 cal/1 g°C ⋅ 25.0 g ⋅ (x - 15.0°C)?

No. (0.5)(25)(0 + 15) + (1)(25)(x - 0)

Chet

 

[Chestermiller]

In post #6, tech99 is discussing a more advanced method of solving a problem like this. It is based on making use of the methodology that you learn in Thermodynamics: since the system is insulated and no work is done by the system, the change in "internal energy" of the system is zero. You will learn all about this when you study Thermo. For now, concentrate on the way that your professor is teaching it to you.

Chet

 

[Eclair_de_XII]

I'll keep that in mind.

No. (0.5)(25)(0 + 15) + (1)(25)(x - 0)

1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = -(0.5 cal/1 g°C ⋅ 25.0 g ⋅ 15.0°C + 80 cal/1 g ⋅ 25.0 g + 1 cal/1 g°C ⋅ 25 g ⋅ x°C)
175x - 15750 = -187.5 - 2000 - 25x
200x = 13562.5
x = 67.8°C

 

[collinsmark]

I'll keep that in mind.
1 cal/1 g°C ⋅ 175.0 g ⋅ (x - 90.0°C) = -(0.5 cal/1 g°C ⋅ 25.0 g ⋅ 15.0°C + 80 cal/1 g ⋅ 25.0 g + 1 cal/1 g°C ⋅ 25 g ⋅ x°C)
175x - 15750 = -187.5 - 2000 - 25x
200x = 13562.5
x = 67.8°C

I think that's right. Good job!

Personally, I would have kept the right hand side of the equation positive and on the left hand side of the equation substituted "x - 90.0^o" with "90.0^o - x" making both sides of the equation represent positive energies. But that's just me. Your approach works too.

 

[Eclair_de_XII]

Thanks again, guys.