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How do you calculate the heat absorbed or released when a substance warms, melts or boils?

Heat and calorimetry: calculate heat changes using q = mC(delta-T) for temperature changes and q = mH for phase changes, with constants from Table B and formulas from Table T.

A focused Regents Chemistry answer on heat and calorimetry: the q = mC(delta-T) equation for warming or cooling, q = mH for melting and boiling, the water constants on Table B, and the difference between exothermic and endothermic changes.

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  1. What this topic is asking
  2. Two equations, two situations
  3. The constants on Table B
  4. Exothermic and endothermic
  5. Try this

What this topic is asking

The Core Curriculum asks you to calculate the heat absorbed or released when a substance changes temperature or changes phase. The Regents gives two equations on Table T, q=mCΔTq = mC\Delta T for temperature changes and q=mHq = mH for phase changes, and the water constants on Table B. This is a reliable Part B-2 and Part C calculation, and it pairs directly with the heating-curve page.

Two equations, two situations

The clue is whether the temperature changes. If the problem gives a starting and ending temperature, you are warming or cooling, so use q=mCΔTq = mC\Delta T. If the problem says "melt", "freeze", "boil" or "condense" at the phase-change point, the temperature is constant, so use q=mHq = mH.

The constants on Table B

Table B lists these for water: specific heat 4.18 J/gK4.18\ \text{J/g}\cdot\text{K}, heat of fusion 334 J/g334\ \text{J/g}, and heat of vaporization 2260 J/g2260\ \text{J/g}. You read them straight off the table rather than memorizing them. Water's high specific heat is why it warms and cools slowly.

Exothermic and endothermic

This links to the kinetics and thermodynamics module, where reaction energy is described the same way. For calorimetry calculations the magnitude of qq is found from the equations, and you state whether the process is exothermic or endothermic from its direction.

Try this

Q1. Calculate the heat released when 10.010.0 g of water cools by 5.05.0 K (specific heat 4.18 J/gK4.18\ \text{J/g}\cdot\text{K}). [2 points]

  • Cue. q=(10.0)(4.18)(5.0)=209q = (10.0)(4.18)(5.0) = 209 J released (exothermic).

Q2. State which equation to use to find the heat needed to boil water at 100 C100\ ^\circ\text{C}. [1 point]

  • Cue. q=mHvq = mH_v (a phase change at constant temperature), using the heat of vaporization from Table B.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (Part C style)3 marksA 50.050.0 g sample of water is heated from 20.0C20.0\,^\circ\text{C} to 60.0C60.0\,^\circ\text{C}. Using the specific heat of water from Table B (4.18 J/gK4.18\ \text{J/g}\cdot\text{K}), (a) show the numerical setup for the heat absorbed and (b) calculate the heat absorbed in joules.
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A 3-point Part C calorimetry calculation using q=mCΔTq = mC\Delta T from Table T and the Table B specific heat.

(a) Setup (1 point): q=mCΔT=(50.0)(4.18)(60.020.0)=(50.0)(4.18)(40.0)q = mC\Delta T = (50.0)(4.18)(60.0 - 20.0) = (50.0)(4.18)(40.0).
(b) Calculation (2 points): (50.0)(4.18)(40.0)=8360(50.0)(4.18)(40.0) = 8360 J (about 8.36×1038.36 \times 10^3 J).

The temperature change is 40.040.0 K (a Celsius change equals a Kelvin change). Markers reward the correct substitution into q=mCΔTq = mC\Delta T and the correct product, with appropriate units of joules.

Regents (Part B-2 style)2 marksDetermine the number of joules required to melt 20.020.0 g of ice at its melting point. (The heat of fusion of water from Table B is 334 J/g334\ \text{J/g}.)
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A 2-point constructed-response item using q=mHfq = mH_f from Table T and the Table B heat of fusion.

Melting at the melting point is a phase change, so use q=mHfq = mH_f rather than q=mCΔTq = mC\Delta T (the temperature does not change). Substitute: q=(20.0 g)(334 J/g)=6680q = (20.0\ \text{g})(334\ \text{J/g}) = 6680 J.

Markers reward selecting the phase-change equation q=mHfq = mH_f (not the temperature-change equation) and the correct product of 66806680 J. Using q=mCΔTq = mC\Delta T here would be wrong because there is no temperature change during melting.

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