At thermal equilibrium, the tea and the ice/water mixture are at the same final temperature.

Since the final temperature is 0.0°C, the tea cools from 20.0°C → 0.0°C, so the tea loses heat:

q_tea = m_tea · c · (ΔT)
where ΔT = T_final − T_initial = 0.0 − 20.0 = −20.0°C, so q_tea is negative (heat released).

That same amount of heat is gained by the ice (law of conservation of energy):

q_tea + q_ice = 0
So, q_ice = −q_tea (this will be positive).

Because the ice starts at 0°C and ends at 0°C, the ice does not warm up. The heat it gains is used only for melting (phase change).

Important point: Final temperature being 0°C does NOT mean all the ice melts.
A mixture of ice + liquid water can remain at 0°C at equilibrium. If the tea does not supply enough heat to melt all the ice, some ice will remain.

Decision check (this is the key step)

Calculate heat released by tea
Use density to convert volume to mass:
m_tea = (volume in mL) × (1.00 g/mL)

Then compute:
q_tea = m_tea · c · (−20.0°C)

Compute heat required to melt all the ice
q_{melt all} = m_ice,total · ΔH_fus

Compare

If q_ice < q_{melt all}, then not all ice melts → final remains 0°C, and some solid ice is left.

If q_ice > q_{melt all}, then all ice melts and the leftover heat warms the water above 0°C (you would then use the remaining heat to find T_final).

If some ice remains (q_ice < q_{melt all})

Use:
q_ice = m_melted · ΔH_fus

So:
m_melted = q_ice / ΔH_fus

Then:
m_{ice remaining} = m_ice,total − m_melted

This question is based on formation enthalpies.
First right the balanced equation
C3H8(g)+5O2(g) →  3CO2(g)+4H2O(g)
Use the formation-enthalpy  equation

here apply the coefficients of the balanced equation
Plug in the values from the given table and you will get ΔH⁰comb in kJ / mol
Answer should be in kJ/g so convert 1mol into grams.
All questions based on formation enthalpies are done by using the equation given above.

I am confused about how to do this kind of question. How do I approach this?

Can you explain to me how to do this kind of question type?

How do I approach this kind of problem?

Hi Mam! I was wondering how to solve this question (with bond enthalpies).

 This is different approach, but the ideas are the same as what we did earlier.

Protonation of the alcohol converts it into a good leaving group.

H₂O leaves, forming a carbocation. This carbocation then undergoes rearrangement to give a more stable carbocation adjacent to the oxygen of the remaining –OH (stabilized by lone pair of oxygen). During this process, ring expansion can occur.

Ring-expansion preference:

If the rearrangement would expand a 6-membered ring to a 7-membered ring, that pathway is generally less favored (more strain / less favorable geometry).

If the rearrangement would expand a 5-membered ring to a 6-membered ring, that pathway is favored because a 6-membered ring is typically more stable than a 5-membered ring.

Method 1 - exocyclic methylene (═CH₂) → endocyclic alkene + methyl substituent (methylene-cycloalkane → methyl-cycloalkene), while retaining the alcohol.

Step 1 — Protect the alcohol (so acid won't dehydrate it)

Reagents: Ac₂O, pyridine (or DMAP)

Intermediate 1: same skeleton, but –OH → –OAc (acetate ester)

Step 2 — Acid-catalyzed alkene isomerization (double-bond migration)

Reagents: cat. p-TsOH (or cat. H₂SO₄), dry solvent (toluene/benzene/CH₂Cl₂), gentle heat
Intermediate 2: isomerized alkene product skeleton, still –OAc
Step 3 — Deprotect (regenerate the alcohol)

Reagents: NaOMe/MeOH or NaOH (aq/alcohol), then workup

Product: target structure with –OH and the shifted alkene + methyl


Method 2 -"hydrogenation → radical bromination → E2 elimination
H₂/Pt hydrogenation converts the exocyclic =CH₂ into a methyl substituent (–CH₃) on that ring carbon.

Br₂, hν does selective radical bromination at the most substituted (tertiary) C–H, which should be the carbon that now bears the methyl group (gives a tertiary bromide).

E2 with EtO⁻ (heat) eliminates HBr to give an alkene. In this fused system, elimination that would put a double bond at a bridgehead is strongly disfavored (Bredt's rule), so the reaction is biased toward the allowed internal alkene you want (in the 5-member ring).



 

I originally thought one product would be result of formation good LG, then E1 elimination to alkene. When tertiary carbocation forms, could other OH group attack electrophile?

Hello! I would like some clarification if possible for this question, thank you so much!