RAC-110 · Lesson 3 of 12 · 25 min

Temperature, sensible heat and latent heat

The brewery's head brewer, Priya, met Dana and Theo at the cooler door with a question. "Our fermenters finish at 18 °C (64 °F), and we crash-cool each batch to 3 °C (37 °F) in the cold room before kegging. On those days the cooler struggles. Is the system too small?" Dana did not answer straight away. She asked how much beer was in a batch, how fast it needed to cool, and whether the brewery ever put ice in the cooler. Then she sat down with Theo and a calculator. "Before we can say whether it is too small," she said, "we have to know how much heat is in that beer."

This lesson covers temperature and its scales, and the two kinds of heat that every refrigeration calculation uses: sensible heat, which changes temperature, and latent heat, which changes state.

Temperature and its scales

Temperature measures how hot or cold something is: in physical terms, the average kinetic energy of its molecules. It is not the same as heat. A bathtub of warm water holds far more heat than a cup of boiling water, but the cup is at the higher temperature. Heat flows from the higher temperature to the lower, whichever body holds more heat in total.

Two everyday scales are used in the trade. The Celsius scale puts the freezing point of water at 0 °C and its boiling point at sea-level pressure at 100 °C. The Fahrenheit scale puts the same two points at 32 °F and 212 °F, so it has 180 Fahrenheit degrees where Celsius has 100, and each Celsius degree is 1.8 Fahrenheit degrees. Two absolute scales start at absolute zero, the temperature at which molecular motion is at its minimum: the kelvin scale uses Celsius-sized degrees and puts absolute zero at 0 K, which is −273.15 °C, and the Rankine scale uses Fahrenheit-sized degrees and puts absolute zero at 0 °R, which is −459.67 °F. Absolute scales are needed in gas laws and radiation.

\[T_{^\circ\text{F}} = 1.8\,T_{^\circ\text{C}} + 32 \qquad T_{^\circ\text{C}} = \frac{T_{^\circ\text{F}} - 32}{1.8} \qquad T_{\text{K}} = T_{^\circ\text{C}} + 273.15 \qquad T_{^\circ\text{R}} = T_{^\circ\text{F}} + 459.67\]

A temperature difference converts differently from a temperature. A difference of 10 Celsius degrees is a difference of 18 Fahrenheit degrees, with no 32 added, because the offset cancels when one temperature is subtracted from another. Superheat, subcooling and the temperature drop across a coil are all differences. Mixing up the two rules is one of the most common conversion errors in the trade.

Worked problem

Converting temperatures and differences

A freezer runs at −23 °C. What is that in °F and in kelvins? The air across its evaporator drops by 6 Celsius degrees. What is that drop in Fahrenheit degrees?

Show solution
  1. \[T_{^\circ\text{F}} = 1.8 \times (-23) + 32 = -9.4\ ^\circ\text{F}\]
  2. \[T_{\text{K}} = -23 + 273.15 = 250.15\ \text{K}\]
  3. \[\Delta T_{^\circ\text{F}} = 1.8 \times 6 = 10.8\ ^\circ\text{F} \quad (\text{no } +32 \text{ for a difference})\]

Answer: −23 °C is about −9.4 °F and 250 K. The 6-degree Celsius drop is a 10.8-degree Fahrenheit drop.

Technicians measure temperature with several kinds of instrument. Thermistors, whose electrical resistance changes with temperature, are used in most digital thermometers, clamp-on pipe probes and the sensors inside modern equipment. Thermocouples, made of two dissimilar metal wires that produce a small voltage, cover a wide range and are used for flue gases and high temperatures. Bimetal and liquid-filled dial thermometers still appear on equipment panels. Infrared thermometers read surface temperature from a distance, with the emissivity caution from Lesson 2. Whatever the instrument, a good technician checks it against a known reference, such as a properly made ice bath at 0 °C (32 °F), before trusting it.

Sensible heat

Sensible heat is heat that changes a substance's temperature without changing its state. It is called sensible because you can sense it with a thermometer. The amount needed depends on the mass, the temperature change and the specific heat, which is the heat needed to raise one unit of mass by one degree.

\[Q = m\,c\,\Delta T\]

In SI units, Q is in kilojoules, m in kilograms, c in kilojoules per kilogram per kelvin and ΔT in kelvins or Celsius degrees. Water has a specific heat of about 4.19 kJ/(kg·K), which is high: water takes a lot of heat to warm and gives a lot back when it cools, which makes it a good medium for carrying heat in chillers and hydronic systems. In imperial units, Q is in Btu, m in pounds, c in Btu per pound per degree Fahrenheit and ΔT in Fahrenheit degrees; the specific heat of water is 1.0 Btu/(lb·°F) by definition of the Btu.

Approximate specific heats, rounded for teaching.
SubstanceApproximate c, kJ/(kg·K)Notes
Water (liquid)4.19High; good heat carrier
Ice2.1About half that of water
Steam or water vapour2.0Varies with pressure and temperature
Beer4.0Slightly less than water
Air (dry)1.006Used in airflow calculations
Aluminum0.90Coil fins
Steel0.49Compressor shells, shelving, carcasses
Copper0.385Tubing

Worked problem

How much heat is in the beer?

A batch is 500 L of beer, about 500 kg, with a specific heat of about 4.0 kJ/(kg·K). It must cool from 18 °C to 3 °C in 6 hours. How much heat must be removed, and at what average rate?

Show solution
  1. \[Q = m\,c\,\Delta T = 500 \times 4.0 \times (18 - 3) = 30\,000\ \text{kJ}\]
  2. \[\dot{Q} = \frac{30\,000\ \text{kJ}}{6 \times 3600\ \text{s}} \approx 1.39\ \text{kW}\]

Answer: About 30,000 kJ, at an average of about 1.4 kW (4,700 Btu/h) over six hours. The cooler's system is rated at 2.6 kW, and the box's other loads take more than half of that, so on crash-cooling days the beer adds a load the system was never sized for. The answer was a glycol jacket on the fermenters, not a bigger cooler.

Product loads like the beer are often the largest part of a refrigeration load in commercial work, and they are easy to forget because they are not part of the building. A grocery delivery of warm produce, a restaurant's hot stock pots put straight into the walk-in, or a load of fresh meat all bring sensible heat that the system must remove. Lesson 11 builds a full load estimate from these pieces.

Latent heat

Latent heat is heat that changes a substance's state without changing its temperature. While ice melts at 0 °C (32 °F), adding heat does not warm the mixture; it loosens the bonds between water molecules until all the ice has turned to water. While water boils at 100 °C (212 °F) at sea level, adding heat does not raise its temperature; it turns liquid into vapour. The word latent means hidden: a thermometer does not show this heat, but it is there, and it is released again when the vapour condenses or the water freezes.

Latent heats are large compared with sensible heats. Melting one kilogram of ice takes about 334 kJ, the same heat that would warm that kilogram of water by about 80 degrees. Boiling one kilogram of water at 100 °C takes about 2,257 kJ, more than five times the heat needed to bring it from freezing to boiling. These are the latent heat of fusion and the latent heat of vaporization of water.

\[Q = m\,h_{fg} \qquad h_{\text{fusion, water}} \approx 334\ \tfrac{\text{kJ}}{\text{kg}}, \qquad h_{fg,\text{water at } 100\,^\circ\text{C}} \approx 2257\ \tfrac{\text{kJ}}{\text{kg}}\]
Heating curve for water: temperature against heat added, rising through ice, flat at the melting point, rising through liquid water, a long flat section at the boiling point, then rising through steam
The heating curve of water. The flat sections are latent heat; the long boiling plateau is why boiling refrigerant can absorb so much heat at one temperature.

Worked problem

From ice to steam

How much heat takes 1 kg of ice at −10 °C to steam at 110 °C (1 lb of ice at 14 °F to steam at 230 °F) at atmospheric pressure? Use the approximate values in the tables.

Show solution
  1. \[Q_1 = 1 \times 2.1 \times 10 = 21\ \text{kJ} \quad \text{(warm the ice)}\]
  2. \[Q_2 = 1 \times 334 = 334\ \text{kJ} \quad \text{(melt it)}\]
  3. \[Q_3 = 1 \times 4.19 \times 100 = 419\ \text{kJ} \quad \text{(warm the water)}\]
  4. \[Q_4 = 1 \times 2257 = 2257\ \text{kJ} \quad \text{(boil it)}\]
  5. \[Q_5 = 1 \times 2.0 \times 10 = 20\ \text{kJ} \quad \text{(superheat the steam)}\]
  6. \[Q = 21 + 334 + 419 + 2257 + 20 = 3051\ \text{kJ}\]

Answer: About 3,050 kJ, and the two latent steps account for 2,591 kJ, or 85 per cent of it. The same pattern holds for refrigerants: most of the heat a refrigerant carries is latent heat.

# Heating curve of water: sensible and latent steps (approximate properties)
c_ice, c_water, c_steam = 2.1, 4.19, 2.0      # kJ/(kg K)
h_fusion, h_vap = 334.0, 2257.0               # kJ/kg
m = 1.0                                       # kg
steps = [
    ("warm ice -10 to 0 C",     m * c_ice * 10),
    ("melt ice at 0 C",         m * h_fusion),
    ("warm water 0 to 100 C",   m * c_water * 100),
    ("boil water at 100 C",     m * h_vap),
    ("superheat steam to 110 C", m * c_steam * 10),
]
total = sum(q for _, q in steps)
running = 0.0
for name, q in steps:
    running += q
    kind = "latent" if "melt" in name or "boil" in name else "sensible"
    print(f"{name:26s} {q:7.0f} kJ  {kind:8s} running total {running:6.0f} kJ")
latent = m * (h_fusion + h_vap)
print(f"latent share of the total: {100 * latent / total:.0f} %")
Output
warm ice -10 to 0 C             21 kJ  sensible running total     21 kJ
melt ice at 0 C                334 kJ  latent   running total    355 kJ
warm water 0 to 100 C          419 kJ  sensible running total    774 kJ
boil water at 100 C           2257 kJ  latent   running total   3031 kJ
superheat steam to 110 C        20 kJ  sensible running total   3051 kJ
latent share of the total: 85 %

Why latent heat runs the refrigeration cycle

A refrigerant is chosen so that it boils at a useful low temperature in the evaporator and condenses at a useful high temperature in the condenser. While it boils, it absorbs its latent heat of vaporization at a nearly constant temperature, which lets a small amount of refrigerant carry a large amount of heat and keeps the evaporator coil at an even temperature from inlet to outlet. In the condenser, the reverse happens: the vapour gives up its latent heat as it condenses. If refrigerants only carried sensible heat, like water in a chiller loop, a system would need to pump far more of them and the coils would have large temperature changes along their length.

Refrigerants have smaller latent heats than water. As approximate teaching values, R-134a absorbs about 200 kJ/kg as it boils at an evaporator temperature near 0 °C, and R-410A about 220 kJ/kg. Water's 2,257 kJ/kg looks far better, but water would have to boil at a deep vacuum to reach refrigeration temperatures, and its vapour takes up an enormous volume at those pressures. Lesson 8 explains how refrigerants are chosen, and Lesson 5 explains why the boiling temperature depends on pressure.

Latent heat also matters on the air side of an air conditioner. Room air contains water vapour. When it crosses an evaporator coil colder than its dew point, some of that vapour condenses on the fins and drips into the drain pan. Condensing that water releases its latent heat into the coil, so the coil must absorb it. Part of an air conditioner's capacity goes to cooling the air, the sensible load, and part goes to removing moisture, the latent load. Lesson 12 returns to this split.

Ice, the original refrigerant

Before mechanical refrigeration, ice was the refrigerant, and it worked because of its latent heat of fusion. An ice box held a block of ice in an insulated compartment, and as heat leaked in, the ice melted at 0 °C (32 °F), holding the box near that temperature until it was gone. The trade's unit of capacity, the ton of refrigeration, comes from this history: it is the cooling effect of melting one short ton (2,000 lb) of ice in 24 hours. Lesson 11 works out its value.

Worked problem

Making ice

The community arena's ice machine in the snack bar makes 100 kg (220 lb) of ice per day from tap water at 15 °C (59 °F). Ignoring the cooling of the ice below the freezing point, what average refrigeration rate does it need just for the water?

Show solution
  1. \[Q_{\text{sensible}} = 100 \times 4.19 \times 15 \approx 6290\ \text{kJ}\]
  2. \[Q_{\text{latent}} = 100 \times 334 = 33\,400\ \text{kJ}\]
  3. \[\dot{Q} = \frac{6290 + 33\,400}{86\,400\ \text{s}} \approx 0.46\ \text{kW}\]

Answer: About 0.46 kW (about 1,570 Btu/h) on average, and 84 per cent of it is the latent heat of freezing. Real machines need more because they also cool the ice below 0 °C, lose heat through the cabinet and spend part of each cycle harvesting.

Common mistakes

  • MistakeConverting a temperature difference with the +32 or −32 offset.

    InsteadDifferences convert by the factor 1.8 alone.

  • MistakeTreating temperature and heat as the same thing.

    InsteadTemperature is a level; heat is an amount of energy. A large cool mass can hold more heat than a small hot one.

  • MistakeForgetting latent heat when estimating a load.

    InsteadFreezing product, condensing moisture and melting ice all involve latent heat, often the largest part.

  • MistakeExpecting a thermometer to rise while ice is melting.

    InsteadDuring a change of state the temperature stays constant while latent heat is absorbed.

Summary and key terms

  • Temperature measures molecular motion; °C and °F are everyday scales, and K and °R are absolute.
  • °F = 1.8 °C + 32 for temperatures, but differences convert by 1.8 alone.
  • Sensible heat changes temperature: Q = mcΔT.
  • Latent heat changes state at constant temperature: Q = m h, with large values for melting and boiling.
  • Refrigerants carry most of their heat as latent heat, boiling in the evaporator and condensing in the condenser.
  • The ton of refrigeration comes from the latent heat of melting ice.
Temperature
A measure of the average kinetic energy of molecules, read on a thermometer.
Absolute zero
The lowest possible temperature, 0 K or −273.15 °C (0 °R or −459.67 °F).
Sensible heat
Heat that changes temperature without a change of state.
Latent heat
Heat absorbed or released during a change of state at constant temperature.
Specific heat
The heat needed to raise a unit mass of a substance by one degree.
Latent heat of fusion
The heat needed to melt a unit mass of a solid at its melting point.
Latent heat of vaporization
The heat needed to boil a unit mass of a liquid at its boiling point.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. Convert 4 °C to °F.

    Show answer

    1.8 × 4 + 32 = 39.2 °F.

  2. How much heat cools 200 kg (440 lb) of water by 10 K (18 °F)?

    Show answer

    200 × 4.19 × 10 = 8,380 kJ.

  3. How much heat freezes 50 kg (110 lb) of water already at the freezing point?

    Show answer

    50 × 334 = 16,700 kJ.

  4. A coil drops air temperature by 11 K (20 °F). What is the drop in Fahrenheit degrees?

    Show answer

    11 × 1.8 = 19.8 °F.

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