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NEET UG Β· Physics

Thermal properties of matter

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NEET UG Β· Physics

Continue studying Thermal Properties Of Matter

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Separate warming from melting, test the final phase and use an energy balance before NEET Thermal Properties of Matter practice. Heat, temperature and phase change: choose the energy steps

About 6 min Β· NEET UG

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Heat, temperature and phase change: choose the energy steps

Separate warming from melting, test the final phase and use an energy balance before NEET Thermal Properties of Matter practice.

What you will understand

For NEET UG Physics students: separate heat from temperature, split warming and phase changes into energy stages, and test whether all the ice can melt before solving for a final temperature.

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Quick answer: identify the phase before the formula

Heat is energy transferred because of a temperature difference; temperature is a state variable, not an amount of stored heat. Within one phase over a range where specific heat c is approximately constant, use Q=mcΞ”TQ=mc\Delta T. During a phase transition of a pure substance at its transition temperature and fixed pressure, use Q=mLQ=mL for the mass changing phase. Its temperature can stay constant while energy enters. A body has internal energy; it does not possess a state-function β€œheat content”.

See where each part of the supplied energy goes

A 0.10 kilogram sample of ice initially at minus 10 degrees Celsius becomes liquid water at plus 20 degrees Celsius at about one atmosphere. Using the supplied constant values ice specific heat 2100 joules per kilogram kelvin, latent heat of fusion 334000 joules per kilogram, and water specific heat 4200 joules per kilogram kelvin: warming the ice to zero takes 2.1 kilojoules; melting all of it at zero takes 33.4 kilojoules without a temperature increase; warming the resulting water to 20 degrees takes 8.4 kilojoules. Total supplied energy is 43.9 kilojoules. The coloured energy bar uses these proportions. Heat losses and container heat capacity are neglected.
Warming ice, melting it and warming the liquid are three different stages. The largest energy share here produces a phase change rather than a temperature rise.

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Ice at -10 Celsius becomes liquid water at 20 Celsius

Take m=0.10 kgm=0.10\,\mathrm{kg} of pure ice at about one atmosphere. Use the supplied constant values cice=2100 J kgβˆ’1 Kβˆ’1c_\mathrm{ice}=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}, Lf=334000 J kgβˆ’1L_f=334000\,\mathrm{J\,kg^{-1}} and cw=4200 J kgβˆ’1 Kβˆ’1c_w=4200\,\mathrm{J\,kg^{-1}\,K^{-1}}. Neglect container heat capacity, evaporation and heat loss. Warm the ice to 0β€‰βˆ˜C0\,{}^\circ\mathrm C: Q1=(0.10)(2100)(10)=2100 JQ_1=(0.10)(2100)(10)=2100\,\mathrm J. Melt all the ice at zero: Q2=(0.10)(334000)=33400 JQ_2=(0.10)(334000)=33400\,\mathrm J. Warm the resulting water to 20: Q3=(0.10)(4200)(20)=8400 JQ_3=(0.10)(4200)(20)=8400\,\mathrm J. Total supplied energy is 43900 J=43.9 kJ43900\,\mathrm J=43.9\,\mathrm{kJ}. Using a single mcΞ”Tmc\Delta T across the entire change would omit latent heat and use the wrong phase’s specific heat.

A calorimetry equation needs a possible final state

For an insulated mixing model with negligible container heat capacity, no evaporation and no other energy input, balance heat lost against heat gained. Before assuming the final mixture is all liquid above zero, compare the heat that warm water can release on cooling to zero with the heat required to melt all the ice at zero. If insufficient, the equilibrium mixture still contains ice at zero under the stated pure-water pressure model. Do not continue an all-liquid equation to an impossible negative temperature while liquid water and ice coexist in this example.

Compare two melting tests before finding a final temperature

Two separate insulated mixing experiments each combine 0.10 kilogram of ice at zero Celsius with 0.20 kilogram of warm water, neglecting evaporation and the cup’s heat capacity. Take water specific heat 4200 joules per kilogram kelvin and ice latent heat of fusion 334000 joules per kilogram near one atmosphere. In A, water initially at 20 Celsius can supply only 16.8 kilojoules on cooling to zero, less than the 33.4 kilojoules needed to melt all the ice. About 50.3 grams melts, leaving 49.7 grams ice and 250.3 grams liquid water at zero. In B, water initially at 60 Celsius can supply 50.4 kilojoules on cooling to zero; after melting, the remaining 17.0 kilojoules warms the total 0.30 kilogram liquid sample to about 13.5 Celsius. These are separate experiments with the stated constant-property model.
The amount of ice is identical in both examples. The initial warm-water temperature changes whether any ice remains at equilibrium.

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Not enough energy to melt all the ice

Mix 0.10 kg0.10\,\mathrm{kg} ice at 0β€‰βˆ˜C0\,{}^\circ\mathrm C with 0.20 kg0.20\,\mathrm{kg} water at 20β€‰βˆ˜C20\,{}^\circ\mathrm C in the stated insulated model. Warm water cooling to zero supplies only (0.20)(4200)(20)=16800 J(0.20)(4200)(20)=16800\,\mathrm J. Melting all the ice would require 33400 J33400\,\mathrm J, so some ice remains. The melted mass is 16800/334000≃0.0503 kg16800/334000\simeq0.0503\,\mathrm{kg}. Final temperature is zero, with about 0.0497 kg0.0497\,\mathrm{kg} ice and 0.2503 kg0.2503\,\mathrm{kg} liquid. The cold material’s phase, not an arithmetic average of the starting temperatures, fixes the final temperature here.

Enough energy to melt the ice and warm the liquid

Repeat with the warm water initially at 60β€‰βˆ˜C60\,{}^\circ\mathrm C. Its potential heat release on cooling to zero is (0.20)(4200)(60)=50400 J(0.20)(4200)(60)=50400\,\mathrm J, greater than the 33400 J33400\,\mathrm J melting requirement. The remaining 17000 J17000\,\mathrm J warms the total 0.30 kg0.30\,\mathrm{kg} liquid sample, giving Tf=17000/[(0.30)(4200)]≃13.5β€‰βˆ˜CT_f=17000/[(0.30)(4200)]\simeq13.5\,{}^\circ\mathrm C. Check the original balance: 0.20cw(60βˆ’Tf)=0.10Lf+0.10cwTf0.20c_w(60-T_f)=0.10L_f+0.10c_wT_f. This checks both the assumed complete melting and the final liquid-water mass.

Temperature differences and free expansion need different checks

A temperature difference of 10 Celsius degrees is 10 K, so a temperature-change calculation does not add 273.15 to Ξ”T\Delta T. Absolute-temperature ratios and radiation formulas require kelvin. For a freely expanding isotropic solid with small changes and a supplied constant linear coefficient, Ξ”L≃αL0Ξ”T\Delta L\simeq\alpha L_0\Delta T. With L0=1.00 mL_0=1.00\,\mathrm m, Ξ±=2.0Γ—10βˆ’5 Kβˆ’1\alpha=2.0\times10^{-5}\,\mathrm{K^{-1}} and Ξ”T=50 K\Delta T=50\,\mathrm K, extension is 0.0010 m=1.0 mm0.0010\,\mathrm m=1.0\,\mathrm{mm}. Restraining expansion can introduce stress; the free-expansion formula alone does not solve that different setup.

Heat amount is different from heat-transfer rate

Specific heat c has unit J kgβˆ’1 Kβˆ’1\mathrm{J\,kg^{-1}\,K^{-1}}; a sample’s heat capacity C is mcmc with unit J/K\mathrm{J/K}. Latent heat L has unit J/kg\mathrm{J/kg}. Thermal conductivity has unit W mβˆ’1 Kβˆ’1\mathrm{W\,m^{-1}\,K^{-1}} and belongs to a heat-flow-rate model, not a phase-change energy. A supplied heater power P gives transferred energy Q=PtQ=Pt only when the specified power actually reaches the sample over time t; account for losses or inefficiency if given.

Common mistakes: name what the energy changes

Do not average temperatures without accounting for masses, heat capacities and phase changes. Do not add 273.15 to a temperature difference. A zero temperature change does not prove zero transferred heat during melting or boiling. Check whether all the ice melts before using the total liquid mass in a final-temperature equation. The approximate constants and pressure conditions in these examples define the model; melting/boiling temperatures and heat capacities are not universal constants independent of conditions.

Try it: melting without a temperature rise

Supply 10.0 kJ10.0\,\mathrm{kJ} to 0.10 kg0.10\,\mathrm{kg} pure ice already at 0β€‰βˆ˜C0\,{}^\circ\mathrm C near one atmosphere, with no losses. Use Lf=334000 J/kgL_f=334000\,\mathrm{J/kg}. Does the sample become warmer liquid water?

Reveal the answer

No. Melting all the ice needs 33.4 kJ33.4\,\mathrm{kJ}, more than the supplied heat. The melted mass is 10000/334000≃0.0299 kg10000/334000\simeq0.0299\,\mathrm{kg}, leaving about 0.0701 kg0.0701\,\mathrm{kg} ice. The mixture remains at zero while both phases coexist under the stated conditions.

Try it: equal specific heat does not mean equal mass

Mix 0.10 kg0.10\,\mathrm{kg} liquid water at 80β€‰βˆ˜C80\,{}^\circ\mathrm C with 0.30 kg0.30\,\mathrm{kg} water at 20β€‰βˆ˜C20\,{}^\circ\mathrm C. No phase change occurs. Neglect heat loss and cup heat capacity; use equal constant specific heats. Is the final temperature 50 Celsius?

Reveal the answer

No. 0.10c(80βˆ’T)=0.30c(Tβˆ’20)0.10c(80-T)=0.30c(T-20) gives T=35β€‰βˆ˜CT=35\,{}^\circ\mathrm C. The simple average of 80 and 20 would apply to equal masses with equal heat capacities per kilogram, not these different masses. Check that the result is between the initial temperatures and closer to the larger cold-water sample’s temperature.

Come back tomorrow: choose the phase before substituting

Explain why melting can absorb heat without a temperature rise. Then predict whether a new mixture has enough energy to melt all its ice before calculating a final temperature. Use the diagram only after making that prediction.

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Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, worked examples and self-checks are original. Supplied property values define simplified teaching models; no measured dataset or human subject review is claimed.

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