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The Hand Warmer Design Challenge

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by

Jason Santana

on 10 December 2013

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Transcript of The Hand Warmer Design Challenge

The Hand Warmer Design Challenge

What substance creates the best handwarmer?
Well then, how will we know which substance will work the best?
1. a). Endothermic

b). The energy released in the formation of hydrated ions is less than the energy required to break the ionic crystal lattice and IMF’s between H20 molecules.

Process
In order to find the change in heat for all of the substances being tested, a variable called the calorimeter constant must be found.
Calculations and some Postlab Questions:

a). qhot=(mhot)(s)( ΔThot)
=(100g)(4.184 J/gC˚)(-20 C˚)
=
-8360J

b). qcold= (mcold)(s)( Δcold)
= (100g)(4.184 J/gC˚)(19 C˚)
=
-7942J

c). qhot= -(qcold+ qcal)
qhot+ qcold= - qcal
-8360J+ -7942J= -qcal

qcal= 688J

d). C= qcal/ΔT
C= 688J/(19C˚) or 688J/292K


C= 36.21 j/C˚ or 2.356 J/K


Jason Santana and Chris Kozak
In order to test which substance will be the most efficient, calorimetry must be used. Calorimetry is the measurement of the amount of heat evolved or absorbed in a chemical reaction, change of state, or formation of a solution.
The substance that, when placed in the calorimeter, gains the most amount of energy and experiences the greatest increase in enthalpy (without increasing too much as to burn a human hand) will be the substance.
Pre-Lab Questions
2). 25 grams of solid A is placed in 60mL of water at an initial temperature at 21.4 degrees Celsius. The final temperature is 25.3 degrees Celsius. Calculate qsolution.

qsolution= -(m×s×ΔT)
qsolution= -(25.0g+60.0g)(4.184 J/gC˚)(25.3 C˚- 21.4 C˚)

qsolution= -1390 J

What is the calorimeter constant?

It is a constant that quantifies the heat capacity of a calorimeter.
How do you find it?
Following the law of conservation of energy, energy can be neither created nor destroyed. That being said, the energies from the reactions inside the calorimeter cannot be either. So first before adding any substances, water will be added into the calorimeter at different temperatures because we know the the heat capacities of water already. One sample of water will be hot while the other will be less warm. When mixed together, equilibrium will be reached by the colder water absorbing energy and the warmer water releasing energy.
Again, following the Law of Conservation of Energy, being the same substance, the colder water should be obtaining the same amount of energy as the hot water is releasing. But by using the equation q= (m×s×ΔT) for each, it is evidently not the case. That's where the calorimeter constant comes in. It accounts for the heat that was lost to the calorimeter itself. So what equation do you use?
Qhot= -(qcold+ qcal)
and
C= qcal/ΔT

So lets begin...
OBSERVATIONS
Based on the results of the lab, hand warmers will benefit most with the use of LiCl since it raised the temperature of the solutions to about 19 degrees Celsius. The only issue with LiCl is that it is very expensive compared to the other materials ($32.75 per 500 grams). The second option however would be MgSO4 that raised the temperature to about 17 degrees, but the price of this substance is only $21.70 per 500 grams?


As a group we decided that Lithium Chloride will be the most efficient substance to use for hand warmers as it will clearly heat up to at last 20 degrees Celsius which, is questions 4, is the situation, although it is expensive.
To complete numbers 2 and 3 use the equations:

qsoln= -(qrxn+CΔT)

qrxn= -(m×s×ΔT)

C is the calorimeter constant
For example:
NaCl
qrxn= -(m×s×ΔT)
qrxn= (5g)(0.864)(-1.2)
qrxn= -5.184

qsoln= -(qrxn+CΔT)
qsoln= -(-5.184+(36.21)(-1.2)
qsoln= 48.636

QUESTION TIME!
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