Question

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/ the styrofoam cup using q = cδt. show your work with units. be sure you have the correct sign on q. (hint: if your water gained heat, 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 surroundings also).
2. what is the total heat gained/lost by the surroundings (water and cup)? be sure you have the correct sign.
3. what is the total heat gained/lost by the system (salt)? be sure you have the correct sign. (hint: heat gained = -heat lost)
4. convert the total heat above into kilojoules. dont forget your sign. this is q.
5. convert the grams of salt into moles. show your work! (molar masses are given on the front page).
6. divide the total heat in kilojoules by the moles of salt. show your work and correct units. (this is enthalpy, δh).

based on the temperature change of your water, was heat gained or lost by the water? the water got hotter 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 always considered to be the surroundings in calorimetry). q = mcδt 49.4 184(27.8 - 22.0) = 979. calculate the amount of heat gained/lost by the water using q = mcδt. show your work with units. be sure you have the correct sign on q.

Explanation:

Step1: Identify the problem type

This is a calorimetry problem involving heat calculations, unit conversions, and enthalpy calculation. The subfield under Natural Science is Chemistry (specifically thermochemistry).

Step2: Analyze each sub - question

Sub - question 4: Heat by surroundings (water + cup)

We use the formula \(q = mc\Delta T\) for water and \(q = C\Delta T\) for the calorimeter (cup). Let's assume we have the mass of water \(m\), specific heat of water \(c = 4.184\ J/g^{\circ}C\), change in temperature \(\Delta T\), and heat capacity of the cup \(C\).
First, calculate \(q_{water}=m\times4.184\ J/g^{\circ}C\times\Delta T\) and \(q_{cup}=C\times\Delta T\). Then \(q_{surroundings}=q_{water} + q_{cup}\). From the given work, if we assume \(m = 49.4\ g\), \(\Delta T=37.8 - 22.0 = 15.8^{\circ}C\) for water and some \(C\Delta T\) for the cup, but the given calculation shows \(49.4\times4.184\times(37.8 - 22.0)=49.4\times4.184\times15.8\approx49.4\times66.1072\approx3266\ J\) (wait, the given work has \(49.4\times4.184\times(37.8 - 22.0)=979\)? Maybe miscalculations in the hand - written part, but the method is \(q = mc\Delta T\) for water and \(q = C\Delta T\) for cup, sum them for surroundings.

Sub - question 5: Heat by system (salt)

Using the hint \(q_{system}=-q_{surroundings}\). So if \(q_{surroundings}\) is calculated as above, \(q_{system}=-q_{surroundings}\).

Sub - question 6: Convert heat to kJ

To convert joules to kilojoules, use the conversion factor \(1\ kJ = 1000\ J\). So if \(q\) is in joules, \(q(kJ)=\frac{q(J)}{1000}\).

Sub - question 7: Convert grams of salt to moles

Use the formula \(n=\frac{m}{M}\), where \(n\) is moles, \(m\) is mass in grams, and \(M\) is molar mass (given on the front page).

Sub - question 8: Calculate enthalpy (\(\Delta H\))

Enthalpy change \(\Delta H=\frac{q_{system}(kJ)}{n}\), where \(n\) is moles of salt from sub - question 7.

(Note: Since the hand - written work has some errors, but the general steps are as above. For example, if we take the given \(q_{surroundings}=979 + 21.76 = 1000.76\ J\) (from the hand - written \(979+21.76 = 1000.76\)), then \(q_{system}=- 1000.76\ J=-1.00076\ kJ\). If we assume mass of salt \(m\) and molar mass \(M\), say \(m = 5\ g\) and \(M = 50\ g/mol\), \(n=\frac{5}{50}=0.1\ mol\), then \(\Delta H=\frac{- 1.00076\ kJ}{0.1\ mol}=-10.0076\ kJ/mol\), but this is just an example with assumed values.)

Answer:

(The answers depend on the actual values of mass, heat capacity, molar mass. But following the steps:

• Sub - question 4: Calculate \(q_{surroundings}=q_{water}+q_{cup}\) using \(q = mc\Delta T\) (water) and \(q = C\Delta T\) (cup).
• Sub - question 5: \(q_{system}=-q_{surroundings}\)
• Sub - question 6: \(q(kJ)=\frac{q(J)}{1000}\)
• Sub - question 7: \(n = \frac{m}{M}\)
• Sub - question 8: \(\Delta H=\frac{q_{system}(kJ)}{n}\)

For the given hand - written calculation (even with errors), if we take \(q_{surroundings}=1000.76\ J\) (from \(979 + 21.76\)), then \(q_{system}=- 1000.76\ J=-1.00076\ kJ\) (sub - question 5), convert to kJ (sub - question 6: \(-1.00076\ kJ\)), then for moles (sub - question 7: need molar mass, say \(M = 58.44\ g/mol\) for NaCl, if \(m = 5\ g\), \(n=\frac{5}{58.44}\approx0.0856\ mol\)), then \(\Delta H=\frac{-1.00076}{0.0856}\approx - 11.7\ kJ/mol\) (approximate, depends on actual values))

Since the problem is a bit unclear with hand - written errors, but the general method is as above for each sub - question.