Example 8.41

Moderators: Chem_Mod, Chem_Admin

Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

Example 8.41

Postby josephyim1L » Sat Feb 09, 2019 9:07 pm

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? Assume that no heat is lost to the surroundings.
can someone explain how to work out this problem...

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: Example 8.41

Postby Matthew Choi 2H » Sat Feb 09, 2019 11:38 pm

The key to this problem is knowing that the heat gained by the water is equal to the heat lost by the water. If you think about what will happen to the system logically, then the ice will melt into water while simultaneously cooling down the water already in the glass. All that's left in the glass will be water that will hypothetically be the exact same temperature. So the final temperature of both the ice and the initial amount of water will be the same. The next step is finding the q for the ice and the water. In order to calculate the q of the ice, you need to use the delta H given and find kJ by converting the 50.0 g into moles and using the stoichiometric ratio. You also need to use (m)(C)(delta T) in order to calculate the heat gained when the water from the melted ice increases in temperature due to it absorbing energy from the initial amount of water. Then, you set that sum equal to the (m)(C)(delta T) of the initial water. For both sides of the equation, you will have one common variable being the final temperature of the water. Then you solve for that variable.

Return to “Phase Changes & Related Calculations”

Who is online

Users browsing this forum: No registered users and 2 guests