Textbook Problem 4C.13

[Julia Mizzi 2I]

Hi! I am having some trouble on problem 4C.13 in the textbook. The problem says "An ice cube of mass 50.0 g at 0°C is added to a glass containing 400.0 g of water at 45°C. What is the final temperature of the system? I am pretty sure I need to add physical change enthalpy somewhere but I'm not sure where. Thanks!

[Sam Leistiko 2B]

You're absolutely right, you do need to add the enthalpy of a physical change! It can be helpful to think about these sorts of problems conceptually before jumping into the math.

If we add ice to water, the water is going to lose heat, and the ice is going to absorb heat. We can assume that all the heat lost by the water is absorbed by the ice. If we look at this problem, we can see that there is a much greater quantity of liquid water than ice. This is because the problem is written to ensure that the ice fully melts, just to make it more simple. You cannot always assume in a problem like this that the ice would fully melt, but in the context of this course, it is (probably) a safe assumption, as it makes the problem less complicated.

Now, lets think about the energy that is transferred. First, the ice needs to melt. All of the heat that the ice absorbs in order to melt comes from the liquid water. Once the ice is melted, the resulting 50.0 g of water needs to heat up until it reaches thermal equilibrium with the other 400.0 g of water. Now, we are ready for the math.

[q][/melting] + [q}[/heating] = - [q][/cooling]
Now, we need to use 1 = mcΔt for each value of q.

So, for q melting, use the mass of the ice as m, and use the heat of fusion of ice as C. You may need to convert to moles. For q heating, use the mass of the ice (which is now liquid water), and use the specific heat of water as c. Finally, for q cooling, use 400.0g as m, because we are talking about the warmer water that was originally in the glass. Once again, use the specific heat of water as C. Finally, this term is negative because we know that energy lost and energy gained have opposite signs, so the amount of energy gained by the ice is -energy lost by the water.

[lanatruong]

Hi, for this problem you are using the equation q (system) = - q (surrounding), with the system being the ice and the surroundings being the water. You have to take into account that the ice has to heat up to become a liquid (phase change) therefore you would use the equation n (in moles) x heat of fusion which would be (2.77542 moles)(6.01 kJ/mol) = 16.68 kJ or 16680 J (convert to joules for later on in the problem). Then you would use the equation
(n x heat of fusion) + (mcdeltaT) = - (mcdeltaT) because you know that q (system) = - q (surrounding). So you would plug in all of the given values you have now.
(16680 J) + (50 g)(4.184 J.C.g)(Tf-0) = - (400g)(4.184 J.C.g)(Tf-45)
You would just isolate and solve for Tf to get your final answer of 31 degrees C. Hope this helps.