Difficulty: Introductory | Prerequisites: Heating and cooling curves, q = mcΔT, states of matter
The heat of fusion is the energy needed to melt a solid at its melting point without changing its temperature. You can measure it experimentally using a calorimeter: add ice to warm water, measure the temperature drop, and use conservation of energy to calculate how much energy the ice absorbed to melt. Many exam problems combine a melting step (using heat of fusion) with a heating step (using q = mcΔT), so you need to be comfortable doing both calculations and adding them together.
Calorimeter
A device used to measure the heat exchanged during a physical or chemical process. In this lab, a simple cup calorimeter holds a known mass of water; when ice is added, the water loses heat and the ice gains it. In simple terms, it is an insulated container that lets you track where the energy goes.
Heat of fusion (Hf)
The energy required to melt one gram of a solid at its melting point, with no change in temperature. For water/ice, the accepted value is approximately 334 J/g. Think of it as the "entry fee" for molecules to break free of the solid crystal structure and become liquid.
Conservation of energy (in calorimetry)
The principle that energy is neither created nor destroyed. In a calorimeter, the heat lost by the warm water equals the heat gained by the ice (assuming no heat escapes to the surroundings). This is the foundation of every calorimetry calculation.
Joule (J) and kilojoule (kJ)
The SI unit of energy. One kilojoule equals 1000 joules. Most calorimetry answers in introductory chemistry land in the range of a few thousand joules to a few tens of thousands, so results are often converted to kilojoules for tidiness.
Mass of ice (from calorimetry data)
Determined indirectly: weigh the calorimeter + water + melted ice at the end, then subtract the mass of calorimeter + water from before the ice was added. The difference is the mass of ice that melted. In simple terms, you figure out how much ice went in by seeing how much heavier everything got.
Temperature change (ΔT)
T_initial minus T_final for the water that lost heat, or T_final minus T_initial for the substance that gained heat. In the heat-of-fusion experiment, the warm water's ΔT is the drop from its starting temperature to the final temperature after the ice melted.
The goal was to measure the heat of fusion of ice experimentally. Warm water was placed in a calorimeter, ice was added, and the final temperature was recorded after the ice melted completely.
Lab data:
Mass of calorimeter: 2.16 g
Mass of calorimeter + water: 105.32 g
Initial water temperature: 25.0 °C
Final water temperature (after ice added): 3.0 °C
Mass of calorimeter + water + melted ice: 137.46 g
Step 1: Find the mass of the water.
Mass of water = (mass of calorimeter + water) − (mass of calorimeter) = 105.32 g − 2.16 g = 103.16 g
Step 2: Find the temperature change of the water.
ΔT = T_initial − T_final = 25.0 °C − 3.0 °C = 22.0 °C
Step 3: Calculate the joules lost by the water.
q = m × c × ΔT = 103.16 g × 4.18 J/g°C × 22.0 °C ≈ 9,486 J (the lab rounded this to 9,500 J)
Step 4: Apply conservation of energy.
The joules needed to melt the ice = the joules lost by the water = 9,500 J. All the heat that left the water went into melting the ice.
Step 5: Find the mass of ice that melted.
Mass of ice = (mass of calorimeter + water + melted ice) − (mass of calorimeter + water) = 137.46 g − 105.32 g = 32.14 g
Step 6: Calculate the heat of fusion.
Heat of fusion = joules to melt ice / mass of ice = 9,500 J / 32.14 g ≈ 296 J/g
The lab worksheet reports 334 J/g (the accepted value). The discrepancy comes from rounding in the joules-lost calculation and heat losses to the surroundings. What matters for the exam is understanding the method and being able to reproduce the calculation.
Q3: 175 g of water heated from 15 °C to 88 °C.
ΔT = 88 − 15 = 73 °C. q = 175 × 4.18 × 73 = 53,399.5 J = 53.40 kJ.
This is a straightforward single-step heating problem. No phase change occurs, so you only need q = mcΔT.
Q4: Energy to melt a 25 g ice cube at 0 °C.
q = mass × heat of fusion = 25 g × 334 J/g = 8,350 J = 8.35 kJ.
No temperature change is involved here; the ice is already at its melting point and stays at 0 °C while it melts. You use the heat of fusion, not q = mcΔT.
Q5: Energy to melt 15 g of ice at 0 °C and then heat the resulting water to 85 °C.
This is a two-step problem:
Step 1, melt the ice: q₁ = 15 g × 334 J/g = 5,010 J = 5.01 kJ
Step 2, heat the liquid water from 0 °C to 85 °C: q₂ = 15 g × 4.18 J/g°C × 85 °C = 5,329.5 J = 5.33 kJ
Total energy: q_total = 5.01 + 5.33 = 10.34 kJ
The key insight is that you must account for both the phase change and the temperature change separately, then add them.
q = m \times c \times \Delta TUse this whenever a substance heats up or cools down without changing state. For water, c = 4.18 J/g°C.
q = m \times H_fWhere Hf is the heat of fusion (334 J/g for ice/water). Use this when ice melts at 0 °C or when any solid melts at its melting point. Temperature does not change during this step.
When a problem asks you to melt ice and then raise the temperature of the resulting liquid:
q_{total} = (m \times H_f) + (m \times c \times \Delta T)First term: energy to melt. Second term: energy to heat the liquid. Add them.
q_{lost\;by\;water} = q_{gained\;by\;ice}The warm water's heat loss equals the ice's heat gain. This relationship lets you solve for any unknown: heat of fusion, mass of ice or final temperature.
To convert joules to kilojoules: divide by 1000.
To convert kilojoules to joules: multiply by 1000.
Students often forget that melting ice at 0 °C and heating liquid water are two separate energy calculations. They use q = mcΔT for everything and skip the heat-of-fusion step. If the problem involves a phase change, you need q = m × Hf as well.
Students sometimes calculate ΔT as a negative number and then get confused about signs. For energy-needed problems at the introductory level, use the absolute value of the temperature change and keep track of direction (lost vs gained) in words, not signs.
A common unit error is forgetting to convert joules to kilojoules (or the reverse) when the question specifies one unit. Always check what the question asks for.
Students sometimes assume that after ice melts, the water is at some temperature above 0 °C. It is not. The water that forms from melting ice is at 0 °C. Any further temperature rise requires additional energy, calculated separately with q = mcΔT.
In the calorimetry experiment, students occasionally mix up which mass to use. The mass in q = mcΔT is the mass of the warm water that lost heat, not the mass of the ice. The mass of ice is used separately when dividing total energy by mass to find the heat of fusion.
Two-step energy problems (melt then heat, or cool then freeze) are a favourite exam question. You will need to calculate each step separately and add the results.
Know the heat of fusion of water (334 J/g) and the specific heat of liquid water (4.18 J/g°C) from memory. These values are often provided on a formula sheet, but having them memorised saves time.
Calorimetry problems test whether you understand conservation of energy: heat lost = heat gained. Expect to be given data for one side and asked to calculate the other.
Be comfortable identifying which formula to use. If the temperature changes, use q = mcΔT. If a phase change occurs at constant temperature, use q = m × Hf (or m × Hv for vaporisation).
Ice, water and steam problems that span multiple regions of a heating curve can appear as a single extended question. Each region gets its own calculation, and the total is the sum of all regions.
True or false: To melt ice at 0 °C, you use the formula q = mcΔT.
False. You use q = m × Hf. There is no temperature change during melting.
Fill in the blank: The heat of fusion of water is approximately ______ J/g.
334
True or false: In a calorimeter, the heat gained by the ice equals the heat lost by the warm water.
True (assuming no heat escapes to the surroundings).
Fill in the blank: To find the mass of ice that melted, you subtract the mass of (calorimeter + water) from the mass of (calorimeter + water + ______).
Melted ice.
True or false: After ice melts at 0 °C, the resulting liquid water is also at 0 °C.
True. Any further warming requires additional energy.
Q: How many kilojoules are required to melt a 25 g ice cube at 0 °C?
A: q = 25 g × 334 J/g = 8,350 J = 8.35 kJ. Only the heat of fusion is needed because the ice is already at its melting point and no temperature change occurs.
Q: How many kilojoules are required to melt 15 g of ice at 0 °C and raise the temperature of the resulting water to 85 °C?
A: Step 1, melt: q₁ = 15 × 334 = 5,010 J = 5.01 kJ. Step 2, heat: q₂ = 15 × 4.18 × 85 = 5,329.5 J = 5.33 kJ. Total = 5.01 + 5.33 = 10.34 kJ.
Q: In a calorimeter, 120 g of water at 30.0 °C cools to 5.0 °C after ice is added. How many joules did the water lose?
A: ΔT = 30.0 − 5.0 = 25.0 °C. q = 120 g × 4.18 J/g°C × 25.0 °C = 12,540 J.
Q: Using the answer above, if 38.0 g of ice melted, what is the experimental heat of fusion?
A: Hf = q / m = 12,540 J / 38.0 g = 330 J/g. This is close to the accepted value of 334 J/g.
Q: Why might a lab measurement of the heat of fusion come out lower than 334 J/g?
A: Heat escapes to the surroundings through the walls of the calorimeter, so the measured heat lost by the water underestimates the true energy absorbed by the ice. This makes the calculated heat of fusion lower than the accepted value.
Heat of fusion ties directly into thermochemistry and enthalpy. In later coursework, you will encounter Hess's law, standard enthalpies of formation and bond energies, all of which build on the same idea: tracking where energy goes during physical and chemical changes. The calorimetry technique used here is the same method used to measure enthalpies of reaction in solution calorimetry experiments.
The conservation-of-energy principle in calorimetry connects to the first law of thermodynamics (energy cannot be created or destroyed). Understanding how to set up "heat lost = heat gained" equations is essential for more advanced topics like enthalpy of neutralisation, enthalpy of combustion and bomb calorimetry.
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