define and use specific heat capacity

Temperature Scales

Learning Objectives

• Understand the Celsius, Fahrenheit and Kelvin scales.
• Convert temperatures between the three scales.
• Define specific heat capacity and use the formula \$Q = mc\Delta T\$.
• Apply specific heat capacity to solve quantitative problems.

Celsius (°C)

The Celsius scale is based on the freezing point (0 °C) and boiling point (100 °C) of water at 1 atm pressure. It is the most commonly used scale in scientific work, except when absolute temperature is required.

Fahrenheit (°F)

The Fahrenheit scale sets the freezing point of water at 32 °F and the boiling point at 212 °F. It is still used in some everyday contexts, particularly in the United States.

Kelvin (K)

Kelvin is the SI unit for thermodynamic temperature. Its zero point (0 K) is absolute zero, the temperature at which molecular motion ceases. The size of one kelvin is identical to one degree Celsius.

Conversion Formulas

All conversions can be derived from the linear relationship between the scales.

• From Celsius to Kelvin: \$K = ^\circ\!C + 273.15\$
• From Kelvin to Celsius: \$^\circ\!C = K - 273.15\$
• From Celsius to Fahrenheit: \$^\circ\!F = \frac{9}{5}\,^\circ\!C + 32\$
• From Fahrenheit to Celsius: \$^\circ\!C = \frac{5}{9}\,(^\circ\!F - 32)\$

Example Conversion

Convert \$25^\circ\!C\$ to Kelvin and Fahrenheit.

1. Kelvin: \$K = 25 + 273.15 = 298.15\ \text{K}\$
2. Fahrenheit: \$^\circ\!F = \frac{9}{5}\times25 + 32 = 45 + 32 = 77^\circ\!F\$

Specific Heat Capacity

Definition

Specific heat capacity (\$c\$) is the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). Its SI unit is \$\text{J kg}^{-1}\text{K}^{-1}\$.

Heat Transfer Equation

\$\$

Q = mc\Delta T

\$\$

where

• \$Q\$ = heat transferred (J)
• \$m\$ = mass of the substance (kg)
• \$c\$ = specific heat capacity (J kg⁻¹ K⁻¹)
• \$\Delta T\$ = change in temperature (K or °C)

Typical Specific Heat Capacities

Worked Example

How much heat is required to raise the temperature of 2.5 kg of water from \$20^\circ\!C\$ to \$80^\circ\!C\$?

1. Identify the data:

   • \$m = 2.5\ \text{kg}\$

   \$c = 4186\ \text{J kg}^{-1}\text{K}^{-1}\$ (water)

   • \$\Delta T = 80 - 20 = 60\ \text{K}\$

2. Apply the formula:

   \$Q = (2.5\ \text{kg})(4186\ \text{J kg}^{-1}\text{K}^{-1})(60\ \text{K})\$

3. Calculate:

   \$Q = 2.5 \times 4186 \times 60 = 627{,}900\ \text{J}\$

4. Result: \$6.28 \times 10^{5}\ \text{J}\$ of heat must be supplied.

Common Mistakes to Avoid

• Mixing temperature units: \$\Delta T\$ must be in kelvin or degrees Celsius, not Fahrenheit.
• Using the specific heat capacity of a different phase (e.g., using water’s \$c\$ for ice).
• For mixtures, ensure each component’s mass and specific heat are accounted for separately.

Practice Questions

1. Convert \$350\ \text{K}\$ to Celsius and Fahrenheit.
2. A 0.75 kg aluminium block is heated from \$25^\circ\!C\$ to \$150^\circ\!C\$. Calculate the heat supplied.
3. Two substances, A (mass 1 kg, \$c = 2000\ \text{J kg}^{-1}\text{K}^{-1}\$) and B (mass 2 kg, \$c = 1000\ \text{J kg}^{-1}\text{K}^{-1}\$), are mixed at \$30^\circ\!C\$ and \$80^\circ\!C\$ respectively. Assuming no heat loss, find the final equilibrium temperature.