Question

based on the temperature change of your water, was heat gained or lost by the water? the water got hotter
calculate the amount of heat gained/lost by the water using ( q = mcdelta t ). show your work with units. be sure you have the correct sign on ( q ). if the temperature of the water went up, ( q ) is + for the surroundings. if the temperature of the water went down, ( q ) is - for the surroundings. water is always considered to be the surroundings in calorimetry. ( q = mcdelta t )

1. using the heat capacity of the calorimeter (a.k.a the cup) calculated in part a (question #4), calculate the amount of heat that was gained/lost by the styrofoam cup using ( q = cdelta t ). show your work with units. be sure you have the correct sign on ( q ). (hint: if your water gained heat, then the calorimeter gained heat, too. if your water lost heat, then the calorimeter lost heat, too. the calorimeter (cup) is considered to be part of the surroundings, too!)

1. what is the total heat gained/lost by the surroundings (water and cup)? be sure you have the correct sign

1. what is the total heat gained/lost by the system (salt)? be sure you have the correct sign. (hint: heat gained = -heat lost)

1. convert the total heat above into kilojoules. dont forget your sign. this is ( q ).

1. convert the grams of salt into moles. show your work! (molar masses are given on the front page)

1. divide the total heat in kilojoules by the moles of salt. show your work and correct units. (this is enthalpy, ( delta h ))

1. fill out the chart below. then, go to your teacher at the computer. your teacher will confirm your math is right.

lab station number
change in temp of water
volume of distilled water used
last answer from part a
answer from #6 above
answer from #8 above

teacher’s approval:

1. add your answer to #8 to the chart at the front of the room. dont forget your sign!

1. copy the chart at the front of the room. circle the salts that are exothermic and therefore could be used to make a handwarmer.

salt	( \text{nh}_4\text{no}_3 )	( \text{cacl}_2 )	( \text{licl} )	( \text{nach}_3\text{co}_2 )	( \text{na}_2\text{co}_3 )	( \text{nacl} )

1. of the circled salts, which two salts would release the most heat upon mixing with water?

1.
2.

1. imagine that you are engineering a hand warmer. you must choose a salt to put in the hand warmer along with a small pouch of water. the costs for all of our salts are listed in the table below. which salt will you choose to put in your hand warmer?

salt	( \text{nh}_4\text{no}_3 )	( \text{cacl}_2 )	( \text{licl} )	( \text{nach}_3\text{co}_2 )	( \text{na}_2\text{co}_3 )	( \text{nacl} )

review questions:

1. when magnesium chloride, ( \text{mgcl}_2 ), is dissolved in water, the temperature of the water drops.

a. is the heat of solution endothermic or exothermic?
b. is the enthalpy of solution positive or negative?

1. a different solution was formed by combining 35.0 g of solid a with 40.0 ml of distilled water, with the water initially at 31.7 °c. the final temperature of the solution was 35.8 °c. calculate the heat released as the solid dissolved, ( q ), assuming no heat loss to the calorimeter. be sure you have the correct sign.
2. in the previous question, the calorimeter was found to have a heat capacity of 5.30 j/°c. calculate the heat of solution, ( q ), including the heat within the calorimeter? be sure you have the correct sign.
3. the solid in question 2 was gallium sulfate, ( \text{ga}_2(\text{so}_4)_3 ). calculate the heat of solution (a.k.a. enthalpy), ( delta h_{\text{soln}} ), for gallium sulfate. check the sign for the enthalpy change!!

Explanation:

Step1: Determine the mass of water

The volume of distilled water is 40.0 mL. The density of water is approximately \(1\space g/mL\), so the mass \(m\) of water is \(m = \text{volume} \times \text{density}=40.0\space mL\times1\space g/mL = 40.0\space g\).

Step2: Calculate the temperature change

The initial temperature \(T_i = 31.7^{\circ}C\) and the final temperature \(T_f = 35.8^{\circ}C\). The temperature change \(\Delta T=T_f - T_i=35.8^{\circ}C - 31.7^{\circ}C = 4.1^{\circ}C\).

Step3: Use the formula \(q = mc\Delta T\) for water

The specific heat capacity of water \(c = 4.184\space J/(g\cdot^{\circ}C)\). Substitute \(m = 40.0\space g\), \(c = 4.184\space J/(g\cdot^{\circ}C)\) and \(\Delta T = 4.1^{\circ}C\) into the formula:
\(q=mc\Delta T=40.0\space g\times4.184\space J/(g\cdot^{\circ}C)\times4.1^{\circ}C\)
First, calculate \(40.0\times4.184 = 167.36\), then \(167.36\times4.1 = 686.176\space J\). Since the temperature of the solution (and water) increased, the water (surroundings) gained heat, so for the system (solid dissolving), the heat released \(q\) (with correct sign) is \(- 686\space J\) (or more precisely \(- 686.176\space J\)). But since the question says "heat released as the solid dissolved", we can report the magnitude with the understanding that the system (solid) released heat, so \(q\approx - 686\space J\) (or in more precise calculation \(q = 40.0\times4.184\times4.1=686.176\space J\) released, so \(q=- 686\space J\) for the system, but the heat gained by surroundings is \(+ 686\space J\); here we follow the hint about correct sign: if temperature went up, surroundings (water) gained heat (\(q\) for surroundings is \(+\)), so for the system (solid), \(q\) is \(-\) as heat gained by surroundings is heat lost by system).

Answer:

The heat released as the solid dissolved is approximately \(\boldsymbol{- 686\space J}\) (or more precisely \(- 686.18\space J\)).