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For question 8.41 I set q(ice)=-q(water) and then used cmdeltaT=-cmdeltaT and solved for Tfinal but i keep getting 42 degrees C instead of 31. Can someone explain what I am doing wrong? The question is A 50.0-g ice cube at 0.0 C is added to a glass containing 400.0 g of water at 45.0 C. What is the final temperature of the system (see Tables 8.2 and 8.3)? Assume that no heat is lost to the surroundings.
Since you are melting the ice cube in water when you place it in the glass, make sure you don't forget to include the enthalpy of fusion term needed to transform an ice cube at 0 degrees into liquid water at 0 degrees. Also make sure that units are consistent (ex. convert J to kJ, or vice versa).
You might also be using the wrong C for the ice. It should be 4.184, the liquid form, because the ice has melted into liquid before it rises in temperature. (I did this and couldn't figure out the problem for a while lol but you might have done what the above person stated, I don't remember the answer I was getting)
ClaireHW wrote:I'd also like to ask if someone could tell us how many steps will have to be calculated and how they come together to get the final temperature.
(Claire Woolson Dis 1K)
The overall goal of the problem is to solve for the final temp. just like in an algebra equation. You'll set up the heat of the water in the ice cube equal to the heat lost by the water in the beaker. To solve for the enthalpy of the water in the ice cube, set up an equation that accounts for the energy of the phase change (solid ice to liquid water) + the energy to raise the liquid water to the final temp. To set up the equation for what's going on with the water in the beaker, use q=mc∆t. Set the two equations equal to each other, but make sure to multiply the water in the beaker equation by a negative sign since heat is being lost. From there, solve for the final temperature.
I hope this helps! :)
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