Calculate the theoretical amount of CaCl2 that would be required to change the remperature of 50ml of water...

Question

Calculate the theoretical amount of CaCl2 that would be required to change the remperature of 50ml of water from room temperature 20-45 degrees celcius

In order to do this calculation you'll need the heat capacity of water (4.184Jg-1deg-1) You can assume that the density of water is 1g/mL
This question is based on the thermochemistry of hot packs!

** Can you please explain step by step how to solve this problem and give the solution as well

thanks

Answer 1

50 ml of water has a mass of 50 g. (Not exactly, but close enough.) It looks like you're saying that room temperature is 20 degrees Celcius (which is accurate), and you want to raise it to 45 degrees, meaning an increase of 25 degrees Celcius. That means the amount of energy you need is (4.184 J/g-C)(50 g)(25 C) = 5,230 J.

There's another piece of information that you need. You need to know how much heat CaCl2 releases when it reacts with water. I found a resource that says a single gram of CaCl2 releases 670 J when it dissolves in water. That means you need to dissolve 5230/670 = 7.81 g of CaCl2. If you have been given a different value for the heat of dissolution of CaCl2, use that instead.

Answer 2

First convert g to moles. 12.00 g x (a million mol ice/18.00 g ice) = 0.6667 mol (notice: Ice is H2O, mw 18.00) heat the forged ice at -5.00oC to 0oC: warmth for temp exchange = mol x (T2-T1) x warmth means 0.6667 mol ice x (0o -(-5o) x 36.5 J/mol oC = 121.7 J or 0.1217 kJ next, soften the ice at 0oC: warmth = mol x C ice 0.6667 mol ice x 6.01 x kJ/mol = 4.007 kJ Then heat the liquid water from 0oC to 0.500oC: 0.6667 mol liquid water x (0.500oC - 0oC) x75.4 J/mol oC = 25.thirteen J or 0.02513 kJ finished warmth: 0.217 kJ +4.007 kJ + 0.0.5 kJ = 4.249 kJ