4C.13

[Alicia Lin 2F]

An ice cube of mass 50.0 g 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 4A.2 and 4C.1)? Assume that no heat is lost to the surroundings.

I went about this problem by using the equation q=mCdeltaT and setting two of these equations equal to each other while using an unknown variable X within the "delta T" and solving for X. I'm not sure if this is somewhat on the right track or not? I am not getting the right answer. Could anyone explain how they went about this?

[Ryan Narisma 4G]

Hello Alicia Lin 2F! First of all, you are correct in setting the equations equal to each other. Did you take into account of the heat required to change the solid water (ice) to liquid water? Because there is an ice cube in the water, reaching the equilibrium temperature requires the heat required to raise the ice cube to the final temperature plus the heat required to change the state from solid @ 0C to liquid @ 0C. I hope this helps!

[Isha_Maniyar_Dis2E]

How did you calculate q for the ice? You should have added the q for the ice undergoing a phase change to the q for the ice melting to become 45 degrees. Then, you set this value to mC(delta t) of the 400.0g of water. Solve for the final temperature, given that the initial is 45 degrees Celsius.

Hope this helped!

[Eileen Si 1G]

Yes, you're on the right track, but for q of ice, you also need to account for the fact that the ice melting into water also gives off energy that needs to be added to the calculation for q=mCdeltaT, so it would be the grams of ice converted into moles and multiplied by the standard enthalpy of physical change of water as given in Table 4C.1 added to q=mCdeltaT.