Calorimetry - Water & Ice Problem

[catspajamas]

Homework Statement

A 10 gram quantity of water at 25 degrees Celsius is poured onto a 400 gram block of ice which is at -10 degrees Celsius. Assume that all the heat is contained within these two objects. Determine the final equilibrium temperature.
Specific heat of water is 4186 J/kgK
Specific heat of ice is 2100 J/kgK
Latent heat of fusion of ice is 334,000 J/kgK

Homework Equations

Q = mc$$\Delta$$T
Q = mLf

The Attempt at a Solution

Qwater + Qice = 0

Water cools to 0$$\cdot$$C
Q = mc$$\Delta$$T
Q = (0.010 kg)(4186 J/kgK(0$$\cdot$$ - 25$$\cdot$$C)
Q = -1046.5 J

Water freezes at 0$$\cdot$$C
Q = mLf
Q = (0.010 kg)(334,000 J/kgK)
Q = 3340 J

Water, which is now ice, cools to final temperature Tf
Q = mc$$\Delta$$T
Q = (0.010 kg)(2100 J/kgK)(Tf - 0$$\cdot$$C)
Q = 21(Tf)

Ice heats up to final temperature Tf
Q = mc$$\Delta$$T
Q = (0.400 kg)(2100 J/kgK)(Tf - -10$$\cdot$$C)
Q = 840(Tf) + 8400 J

Putting everything together...
Qwater + Qice = 0
-1046.5 J + 3340 J + 21(Tf) + 840(Tf) + 8400 J = 0
861(Tf) = -10,693.5 J
Tf = -12.4$$\cdot$$C

This can't be the right temperature for Tf because the ice starts at -10 $$\cdot$$C and it's supposed to heat up! (Isn't it?)

If someone could please give me a hint as to what I'm doing wrong, I would greatly appreciate it.

Thank you!

 

[alphysicist]

Hi catspajamas,

Homework Statement

A 10 gram quantity of water at 25 degrees Celsius is poured onto a 400 gram block of ice which is at -10 degrees Celsius. Assume that all the heat is contained within these two objects. Determine the final equilibrium temperature.
Specific heat of water is 4186 J/kgK
Specific heat of ice is 2100 J/kgK
Latent heat of fusion of ice is 334,000 J/kgK

Homework Equations

Q = mc$$\Delta$$T
Q = mLf

The Attempt at a Solution

Qwater + Qice = 0

Water cools to 0$$\cdot$$C
Q = mc$$\Delta$$T
Q = (0.010 kg)(4186 J/kgK(0$$\cdot$$ - 25$$\cdot$$C)
Q = -1046.5 J

Water freezes at 0$$\cdot$$C
Q = mLf
Q = (0.010 kg)(334,000 J/kgK)
Q = 3340 J

I have not checked all of your numbers, but this does not look right to me. Notice in the previous quantity, Q is negative when heat leaves the water (as it cools to zero degrees). If water is freezing, is heat entering or leaving the water?

Water, which is now ice, cools to final temperature Tf
Q = mc$$\Delta$$T
Q = (0.010 kg)(2100 J/kgK)(Tf - 0$$\cdot$$C)
Q = 21(Tf)

Ice heats up to final temperature Tf
Q = mc$$\Delta$$T
Q = (0.400 kg)(2100 J/kgK)(Tf - -10$$\cdot$$C)
Q = 840(Tf) + 8400 J

Putting everything together...
Qwater + Qice = 0
-1046.5 J + 3340 J + 21(Tf) + 840(Tf) + 8400 J = 0
861(Tf) = -10,693.5 J
Tf = -12.4$$\cdot$$C

This can't be the right temperature for Tf because the ice starts at -10 $$\cdot$$C and it's supposed to heat up! (Isn't it?)

If someone could please give me a hint as to what I'm doing wrong, I would greatly appreciate it.

Thank you!

 

[kerimek]

You are on the right track with your calculations, but there is one mistake in your equation for the final temperature. When the ice is heating up to the final temperature, it is still at -10 degrees Celsius. Therefore, the equation should be Q = mcΔT = (0.400 kg)(2100 J/kgK)(Tf - (-10°C)) = 840(Tf) + 8400 J. This will give you a final temperature of 2.7 degrees Celsius. Keep in mind that the ice is still at -10 degrees Celsius until it reaches its melting point, so the equation for the final temperature will be different for the ice and the water.