9.5 Specific Heat and Thermal Conductivity

Specific Heat and Energy Required to Change Temperature

Specific heat is a fundamental property that determines how much energy is needed to change an object's temperature. It varies widely between materials and explains why some substances heat up quickly while others take much longer.

When thermal energy is added to an object, its temperature changes according to the relationship:

$Q = mc\Delta T$

Where:

• $Q$ = heat energy transferred (J)
• $m$ = mass of the object (kg)
• $c$ = specific heat of the material (J/kg·°C)
• $\Delta T$ = change in temperature (°C)

This equation works in both directions - it can determine how much energy is needed to change a temperature, or how much temperature will change when energy is added.

Specific heat is an intrinsic property unique to each material:

• Water has an exceptionally high specific heat (4,186 J/kg·°C), which is why it takes significant energy to heat water
• Metals like aluminum (900 J/kg·°C) and copper (385 J/kg·°C) have lower specific heats, explaining why they heat up more rapidly
• The specific heat of a material is determined by its molecular structure and the types of bonds between atoms

Thermal Conductivity and Rate of Energy Transfer

Thermal conductivity measures how efficiently a material transfers heat energy. When there's a temperature difference across a material, heat flows from the hotter region to the cooler region at a rate determined by several factors.

The rate of heat transfer through conduction is given by:

$\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$

Where:

• $\frac{Q}{\Delta t}$ = rate of heat transfer (W or J/s)
• $k$ = thermal conductivity (W/m·°C)
• $A$ = cross-sectional area (m²)
• $\Delta T$ = temperature difference (°C)
• $L$ = thickness of the material (m)

This equation shows that heat transfers more quickly when:

• The material has higher thermal conductivity
• The cross-sectional area is larger
• The temperature difference is greater
• The material is thinner

Thermal conductivity varies dramatically between materials:

• Metals are excellent conductors with high values (copper: 401 W/m·°C, aluminum: 205 W/m·°C)
• Insulators have very low thermal conductivity (wood: 0.12 W/m·°C, fiberglass: 0.04 W/m·°C)
• This property explains why metal feels cold to the touch - it rapidly conducts heat away from your hand

The thermal conductivity of a material is determined by its atomic structure and how freely electrons can move within it, which is why metals (with their free electrons) are typically good thermal conductors.

Practice Problem 1: Specific Heat

Solution

First, we need to use the equation $Q = mc\Delta T$ for both the aluminum pot and the water.

For the entire system: $Q_{total} = Q_{aluminum} + Q_{water}$ $84,000 \text{ J} = m_{al}c_{al}\Delta T + m_{water}c_{water}\Delta T$ $84,000 \text{ J} = (2.0 \text{ kg})(900 \text{J/kg} \cdot \text{°C})\Delta T + (0.5 \text{ kg})(4,186 \text{J/kg} \cdot \text{°C})\Delta T$ $84,000 \text{ J} = (1,800 \text{ J/°C} + 2,093 \text{ J/°C})\Delta T$ $84,000 \text{ J} = 3,893 \text{ J/°C} \times \Delta T$ $\Delta T = \frac{84,000 \text{ J}}{3,893 \text{ J/°C}} = 21.6°\text{C}$

Therefore, the final temperature is $20°\text{C} + 21.6°\text{C} = 41.6°\text{C}$.

Practice Problem 2: Thermal Conductivity

Solution

We can use the thermal conductivity equation to find the rate of heat transfer:

$\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$

Where:

• $k = 0.8 \text{W/m} \cdot \text{°C}$ (thermal conductivity of glass)
• $A = 1.5 \text{ m} \times 2.0 \text{ m} = 3.0 \text{ m}^2$ (area of the window)
• $\Delta T = 22°\text{C} - (-5°\text{C}) = 27°\text{C}$ (temperature difference)
• $L = 0.006 \text{ m}$ (thickness of the glass)

Substituting these values:

$\frac{Q}{\Delta t} = \frac{(0.8 \text{W/m} \cdot \text{°C})(3.0 \text{ m}^2)(27°\text{C})}{0.006 \text{ m}}$ $\frac{Q}{\Delta t} = \frac{64.8 \text{ W·m}}{0.006 \text{ m}} = 10,800 \text{ W}$

Therefore, the rate of heat loss through the window is 10,800 watts or 10.8 kilowatts.