Specific heat capacity and specific latent heat in calculations
Calculate warming stages with Q = mcΔT and melting or boiling stages with Q = mL. For a pure substance at constant pressure, temperature stays constant during melting or boiling. Add stage energies and distinguish heater output from energy reaching the sample.
What a learner can do afterwards
- Works a multi-stage problem that both warms and melts a sample
- Reads a temperature-time graph and names the flat sections as changes of state
- Uses a heater's power and running time as the energy supplied, and comments on losses to the surroundings
1 · Read
You have used Q = mcΔT for warming without a state change. Q is energy absorbed in J, m is mass in kg, c is in J/(kg°C), and ΔT is final minus initial temperature in °C. Divide grams by 1000 to get kg. Use c for the current state: about 2100 for ice and 4200 for liquid water. Warming ice from -10°C to 0°C gives a rise of 10°C.
For melting or boiling, use Q = mL. Specific latent heat, L, is energy per kilogram for that state change, in J/kg. Choose the melting or boiling value as needed. Pure water melts at about 0°C and boils at about 100°C at normal atmospheric pressure. During either change at constant pressure, temperature stays constant. Energy changes particle arrangements and interactions, not the chemical bonds within water molecules.
Take 0.1 kg of ice at -5°C to water at 20°C at normal atmospheric pressure. Use c_ice = 2100 J/(kg°C), L = 334000 J/kg and c_water = 4200 J/(kg°C). Warm ice: 0.1 x 2100 x 5 = 1050 J. Melt: 0.1 x 334000 = 33400 J. Warm water: 0.1 x 4200 x 20 = 8400 J. Add them: 1050 + 33400 + 8400 = 42850 J absorbed. Ignore transfers elsewhere.
Heater output is E = Pt: power in W times time in s gives energy in J. One watt is one joule per second. A 60 W heater running for 90 s supplies 5400 J. If 1000 J warms the container and surroundings, the sample receives 4400 J. Subtract stated transfers elsewhere before using sample energy in mcΔT or mL. To reach the same final state with losses, the heater must supply more energy.
A heating curve plots temperature vertically against time horizontally. With positive net energy entering a pure sample at constant pressure, rising sections show warming and horizontal sections show melting or boiling. For water at normal atmospheric pressure, the 0°C plateau shows melting and the 100°C plateau shows boiling. A flat trace alone does not prove a state change: temperature can also stay steady when no net energy enters.
If only part of the ice melts, subtract any earlier warming energy first. The energy left for melting gives melted mass = Q/L. For example, 3340 J available for melting with L = 334000 J/kg melts 0.01 kg. Compare this with the starting mass to check whether ice remains. During melting at constant pressure, ΔT = 0 does not mean no energy is needed: mL is positive for a positive melted mass.
Use mcΔT for warming and mL for melting or boiling. Add stage energies. Heater output is Pt; subtract transfers elsewhere to find energy reaching the sample.
Lesson, questions and flashcards written for this dataset. OpenStax textbook records were consulted for scientific facts and problem structures; their prose was not reproduced.
Readings, questions and flashcards written for this dataset. OpenStax textbook records were studied for scientific facts and problem structures, not copied or closely paraphrased.
More to read
Plan the stages before calculating
Write down the starting state and temperature, then the finishing state and temperature. One temperature difference can hide a melting stage that still needs energy.
For ice below 0°C becoming water above 0°C at normal atmospheric pressure, first warm the ice to its melting point. Use the specific heat capacity of ice. Next calculate melting energy with mL. Finally warm the liquid using the specific heat capacity of water.
Keep the three energies separate until the last addition. Their sum is the energy absorbed by the sample. If the container and surroundings also receive energy from the heater, its required output is larger.
A calculation that stops after melting leaves water at the melting temperature. It has not yet accounted for warming that water to a higher finishing temperature.
Calculate how much ice melts
A heater may stop before all the ice has melted. First account for any energy used to bring colder ice to its melting point. Subtract that warming energy from the energy absorbed by the sample.
The remaining energy can melt ice. Rearrange Q = mL to melted mass = Q/L. In this calculation, m means the mass that melts, not automatically the starting mass.
For example, with L = 334000 J/kg, 6680 J available for melting melts 0.020 kg. If the sample contained more ice, some remains. At normal atmospheric pressure, ice and liquid water can remain together at about 0°C.
Check the result against the starting mass. If the calculated melted mass exceeds the available ice, the energy account needs to include what happens after melting finishes.
Read the trace and the experiment together
On a temperature-time graph, a rising section records increasing temperature. A horizontal section records a period with no temperature change. The graph does not tell you by itself where energy is going.
For the usual heating-curve model, the sample is pure, pressure stays constant and positive net energy enters. A horizontal section then shows melting or boiling. Energy changes the state while the temperature stays at the transition temperature.
At normal atmospheric pressure, pure water has a melting plateau at about 0°C and a boiling plateau at about 100°C. These temperatures identify the stages only when the experimental conditions support that interpretation.
A sample receiving no net energy can also have a horizontal trace. Check the heater and energy-transfer information before naming a state change.
2 · Watch
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