Pizza Rolls 3B

Moderators: Chem_Mod, Chem_Admin

Jason Wu 1E
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

Pizza Rolls 3B

Postby Jason Wu 1E » Tue Feb 11, 2020 9:30 pm

Dr. Lavelle picks up the ice cream that he has just heated up (-2.8 Celsius) and accidentally drops it on the ground and can’t eat it. Tears streaming down his face, he watches as half of it melts away when he realizes that from the moment he dropped the ice cream until now, it has received exactly 234kJ of heat. After performing quick calculations, he realizes that the enthalpy of fusion (kJ/g) for the ice cream is?

I am lost on how to start this problem and solving it. Also, unsure about if this relates? But does taking the area under the constant temperature of the heating curve give the enthalpy of fusion/vaporization?

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: Pizza Rolls 3B

Postby nicolely2F » Tue Feb 11, 2020 9:51 pm

The ice cream receives a total of 234 kJ. Part of it is used to increase its temperature from -2.8 C to fusion point at 0 C, the rest is used in the phase change of half of the mass of the ice cream (that is, 125 g / 2 = 62.5 g) from solid to liquid.

First equation (temperature change): q = m.c.∆T = (125)(0.010)(2.8) = 3.5 kJ
> The value for c, 0.010, was discovered in part A

Second equation (phase change): ∆H (fusion) = q / m = (234 - 3.5) / (62.5) = 3.7 kJ / g

Cooper Baddley 1F
Posts: 100
Joined: Wed Sep 18, 2019 12:19 am

Re: Pizza Rolls 3B

Postby Cooper Baddley 1F » Tue Feb 11, 2020 10:09 pm

The formula for the phase change is Energy = mass*entalpy of fusion. You are given that half of the original 125g melts so you have 62.5g and then that the total energy is 234 KJ. Plug those in and solve for enthalpy of fusion.


Return to “Heat Capacities, Calorimeters & Calorimetry Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest