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Interactive calorimetry & thermal inertia laboratory

Specific Heat Capacity (c = Q / mΔT)

Investigate how substances store thermal energy. Explore calorimetry mixtures by dropping heated metal blocks into insulated liquid cups, or run side-by-side electric heating trials to witness thermal inertia in action.

Specific Heat Capacity Laboratory

Select a tab above. Drop heated metal blocks (Aluminum, Copper, Iron, Lead, Unknown) into beakers of water, alcohol, or oil. Or heat liquids side-by-side with immersion heaters to view temperature curves over time.

Ready

Live Lab Telemetry

Liquid Temp (\\(T_L\\))
20.0 °C
Metal Temp (\\(T_M\\))
100.0 °C
Equilibrium (\\(T_f\\))
Pending
Heat Lost (\\(Q_{\\text{lost}}\\))
0.0 J
Heat Gained (\\(Q_{\\text{gain}}\\))
0.0 J
Liquid A Temp (\\(T_A\\))
20.0 °C
Liquid B Temp (\\(T_B\\))
20.0 °C
Energy Supplied
0 J
Heating Time
0 s

What is Specific Heat Capacity?

Specific Heat Capacity (symbolized by c) is a physical property of matter defined as the quantity of heat energy required to raise the temperature of one kilogram of a substance by one Kelvin (or one degree Celsius).

This quantity measures the "thermal inertia" of a material. Materials with a high specific heat capacity (like water) require a large amount of energy to change their temperature, which makes them excellent heat reservoirs. Conversely, materials with low specific heats (like metals) experience rapid temperature changes when heated.

Key Definitions

  • Specific Heat Capacity (c): The energy needed to heat 1 kg of a material by 1 K. Unit: J/(kg·K) or J/(kg·°C).
  • Heat Capacity (C): The energy needed to heat a whole object by 1 K. Unit: J/K. Related by C = m * c.
  • Thermal Equilibrium: The state where two objects in contact reach the same temperature, ending net heat flow.

Heat Formulas

Sensible Heat Equation Q = m * c * ΔT
Change in Temp ΔT = Tfinal - Tinitial
Calorimetry Mixture m1c1(T1-Tf) = m2c2(Tf-T2)
Variables and Units
  • Q = heat transferred (Joules, J)
  • m = mass of substance (kilograms, kg)
  • c = specific heat capacity (J/(kg·K))
  • ΔT = temperature change (°C or K)

Specific Heats of Common Substances

Substance State c (J/(kg·K)) c (cal/(g·°C))
Water Liquid 4,184 1.000
Ethanol Liquid 2,440 0.583
Glycerin Liquid 2,430 0.580
Olive Oil Liquid 2,000 0.478
Ice (0°C) Solid 2,100 0.502
Steam (100°C) Gas 1,996 0.477
Aluminum Solid 900 0.215
Iron / Steel Solid 450 0.108
Copper Solid 385 0.092
Lead Solid 128 0.031

Solved Examples

A 0.5 kg aluminum block (c = 900 J/(kg·K)) at 100°C is dropped into a container holding 1.0 kg of water (c = 4184 J/(kg·K)) at 20°C. Calculate the final equilibrium temperature, assuming no heat loss to the surroundings.
  1. Identify the given values: mass of block (m_m) = 0.5 kg, specific heat of aluminum (c_m) = 900 J/(kg·K), initial temperature of metal (T_m) = 100°C. Mass of water (m_w) = 1.0 kg, specific heat of water (c_w) = 4184 J/(kg·K), initial temperature of water (T_w) = 20°C.
  2. Use the heat conservation equation: Q_lost = Q_gained.
  3. Substitute variables: m_m * c_m * (T_m - T_f) = m_w * c_w * (T_f - T_w).
  4. Plug in the numbers: 0.5 * 900 * (100 - T_f) = 1.0 * 4184 * (T_f - 20).
  5. Simplify: 450 * (100 - T_f) = 4184 * (T_f - 20) => 45000 - 450 T_f = 4184 T_f - 83680.
  6. Combine like terms: 45000 + 83680 = (4184 + 450) * T_f => 128680 = 4634 * T_f.
  7. Solve for T_f: T_f = 128680 / 4634 ≈ 27.77°C.
  8. Verify: The final temperature is between 20°C and 100°C, and is closer to 20°C because water has a much larger thermal inertia (heat capacity) than aluminum.

Answer: T_f = 27.8°C

An electric immersion heater rated at 250 W is used to heat 0.4 kg of olive oil (c = 2000 J/(kg·K)) in an insulated beaker. If the oil starts at 15°C, how long will it take to heat the oil to 75°C? Neglect the heat capacity of the beaker.
  1. Identify the given values: power (P) = 250 W = 250 J/s, mass (m) = 0.4 kg, specific heat (c) = 2000 J/(kg·K), temperature rise (ΔT) = 75°C - 15°C = 60°C (or 60 K).
  2. Calculate the heat energy required (Q) using: Q = m * c * ΔT.
  3. Substitute the values: Q = 0.4 * 2000 * 60 = 48,000 Joules.
  4. Recall the relation between energy and power: Q = P * t, where t is time in seconds.
  5. Solve for time (t): t = Q / P = 48,000 / 250 = 192 seconds.
  6. Convert to minutes: 192 s = 3 minutes and 12 seconds.
  7. Verify: 192 s at 250 W delivers 48 kJ of energy, which exactly heats 0.4 kg of oil by 60°C.

Answer: Time = 192 seconds (3m 12s)

A 120 g copper block at 95°C is dropped into 180 g of an unknown liquid at 18°C. The system reaches an equilibrium temperature of 23.5°C. If copper has c = 385 J/(kg·K), calculate the specific heat capacity of the unknown liquid.
  1. Identify the given values: mass of copper (m_c) = 120 g = 0.12 kg, specific heat of copper (c_c) = 385 J/(kg·K), initial temperature of copper (T_c) = 95°C. Mass of liquid (m_L) = 180 g = 0.18 kg, initial temperature of liquid (T_L) = 18°C. Equilibrium temperature (T_f) = 23.5°C.
  2. Calculate heat lost by the copper block: Q_lost = m_c * c_c * (T_c - T_f) = 0.12 * 385 * (95 - 23.5) = 46.2 * 71.5 = 3,303.3 Joules.
  3. Use conservation of energy: Q_gained by liquid = Q_lost = 3,303.3 Joules.
  4. Recall the heat gained equation: Q_gained = m_L * c_L * (T_f - T_L).
  5. Substitute values: 3,303.3 = 0.18 * c_L * (23.5 - 18) => 3,303.3 = 0.18 * c_L * 5.5 = 0.99 * c_L.
  6. Solve for c_L: c_L = 3,303.3 / 0.99 = 3,336.67 J/(kg·K).
  7. Verify: The specific heat capacity is around 3337 J/(kg·K), which is higher than typical organic liquids like ethanol (2440) but lower than water (4184), a reasonable value for aqueous mixtures.

Answer: c_liquid = 3,337 J/(kg·K)

Common Mistakes

  • Confusing Specific Heat with Heat Capacity: Remember that specific heat (c) is independent of the size of the sample, while heat capacity (C = m·c) depends on the object's mass.
  • Mixing Units: Mixing grams and kilograms, or Joules and calories. Always convert mass to kg if using c in J/(kg·K).
  • Sign errors in ΔT: ΔT is T_final - T_initial. When calculating Q_lost, it is convenient to write m·c·(T_initial - T_final) to keep it positive, but make sure the equations match.
  • Assuming Specific Heat is constant through phase changes: Solid ice (2100 J/kg·K) and liquid water (4184 J/kg·K) have very different specific heats.

Practice Questions

1. Why does a sandy beach feel scorching hot in the afternoon while the ocean water remains pleasantly cool, even though both receive the same amount of solar radiation?

This is due to the difference in their specific heat capacities. Dry sand has a low specific heat capacity (around 800 J/(kg·K)), meaning it requires very little heat energy to raise its temperature. Water has a very high specific heat capacity (4184 J/(kg·K)), meaning it can absorb a massive amount of solar heat with only a minor rise in temperature. Thus, sand heats up much faster and hotter than water during the day.

2. Why is water widely used as the cooling fluid in automotive radiators and industrial heat exchangers?

Water possesses an exceptionally high specific heat capacity (4184 J/(kg·K)). This allows water to absorb enormous quantities of waste heat from engines or machinery while undergoing relatively small temperature changes. This prevents the coolant from boiling easily and makes heat transport highly efficient.

3. Explain the difference between Specific Heat Capacity (c) and Heat Capacity (C). How are they related mathematically?

Specific Heat Capacity (c) is an intrinsic property of a substance representing the heat needed to raise the temperature of 1 kg of the material by 1°C (units: J/(kg·K)). Heat Capacity (C) is an extrinsic property representing the heat needed to raise the temperature of a specific object by 1°C (units: J/K). They are related by mass (m): C = m * c.

4. A copper frying pan and an iron frying pan of equal mass are placed on identical burners. Which pan will heat up faster? Explain using their specific heat values.

Copper has a lower specific heat capacity (385 J/(kg·K)) than iron (450 J/(kg·K)). Since c_copper < c_iron, the copper pan requires less thermal energy input per kilogram to raise its temperature by 1°C. Therefore, under identical heat input rates, the copper pan will increase in temperature faster than the iron pan.

FAQ

Frequently Asked Questions

What is Specific Heat Capacity?

Specific heat capacity (symbolized as c) is the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin (or 1°C). Its SI unit is Joules per kilogram Kelvin, J/(kg·K).

What is the formula for Specific Heat Capacity?

The heat energy formula is Q = m * c * ΔT. Rearranging to solve for specific heat capacity gives: c = Q / (m * ΔT), where Q is heat in Joules, m is mass in kg, and ΔT is the change in temperature in °C or K.

Why does water have such a high specific heat capacity?

Water has a high specific heat capacity because of hydrogen bonding. When heat is added, much of the energy is first used to break the strong intermolecular hydrogen bonds between water molecules rather than increasing their kinetic energy (which determines temperature).

What is the difference between specific heat capacity and specific heat?

In practice, they are often used interchangeably. Technically, "specific heat" is a dimensionless ratio comparing the heat capacity of a substance to that of water, whereas "specific heat capacity" includes physical units (J/(kg·K)).

How is the equilibrium temperature calculated in a mixture?

For an insulated mixture of two substances, the heat lost by the hot substance equals the heat gained by the cold substance: m_1 * c_1 * (T_1 - T_f) = m_2 * c_2 * (T_f - T_2). Solving this equation yields the final temperature: T_f = (m_1 c_1 T_1 + m_2 c_2 T_2) / (m_1 c_1 + m_2 c_2).

TemperatureHeatThermal EnergyCelsius ScaleSpecific Heat CapacityLatent HeatFirst Law of Thermodynamics