College Physics - Chapter 14: Heat and Heat Transfer Methods

Heat and Heat Transfer Methods

Introduction to Heat

• Heat transfer occurs due to temperature differences between objects or systems.
• Heat is defined as energy that is transferred solely because of a temperature difference.
• When objects with different temperatures interact (e.g., a soft drink and ice), energy is transferred until they reach thermal equilibrium at the same temperature ($T'$).

Mechanical Equivalent of Heat

• It's possible to alter a substance's temperature by performing work on it.
• This demonstrates that heat is a form of energy.
• James Prescott Joule's experiments established the mechanical equivalent of heat, which relates work to heat transfer.
• $1.00 \text{ kcal} = 4186 \text{ J}$

Heat Transfer and Temperature Change

• The amount of heat transferred is directly proportional to the temperature change.
  • Doubling the temperature change requires twice the heat.
• The amount of heat transferred is directly proportional to the mass.
  • Doubling the mass requires twice the heat for an equivalent temperature change.
• The amount of heat transferred depends on the substance and its phase.
  • Different substances require different amounts of heat to achieve the same temperature change.

Factors Affecting Heat Transfer

• Transferred heat ($Q$) depends on:
  • Change in temperature ($\Delta T$)
  • Mass of the system ($m$)
  • Substance and its phase
• Equation for heat transfer: $Q = mc\Delta T$
  • $Q$ = heat transfer
  • $m$ = mass
  • $\Delta T$ = change in temperature
  • $c$ = specific heat (depends on the material and phase)
• SI unit for specific heat: J/(kg·K) or J/(kg·°C)
• Specific heat is the amount of heat needed to change the temperature of 1.00 kg of a substance by 1.00 °C.

Example: Heating Water in an Aluminum Pan

• A 0.500 kg aluminum pan is used to heat 0.250 kg of water from 20.0 °C to 80.0 °C.
• Specific heat of water: 4186 J/kg·°C
• Specific heat of aluminum: 900 J/kg·°C
• The pan and water are at the same temperature.
• When the pan is heated, both the pan and water experience the same temperature increase.

Phase Change and Latent Heat

• During a phase change (e.g., ice melting), no temperature change occurs despite heat transfer.
• Most substances exist in solid, liquid, or gas phases.
• Phase changes occur at fixed temperatures for a given substance and pressure (boiling and freezing/melting points).
• Heat absorbed or released during phase changes is given by:
  • $Q = mL_f$ (melting/freezing)
  • $Q = mL_v$ (vaporization/condensation)
  • $L_f$ = latent heat of fusion
  • $L_v$ = latent heat of vaporization

Example: Chilling Soda with ice cubes

• Three ice cubes at 0 °C are used to chill 0.25 kg of soda at 20 °C.
• Each ice cube has a mass of 6.0 g.
• The soda is in a foam container to minimize heat loss.
• Assume the soda has the same heat capacity as water (4186 J/kg·°C).
• Latent heat of fusion of water: 334 kJ/kg
• Melting of ice occurs in two steps: phase change (ice to water at 0°C), then temperature increase of the water.

Heat Transfer Methods

• Heat is transferred by conduction, convection, and radiation.
• In a fireplace, radiation is primarily responsible for heat transfer.
• Conduction occurs at a slower rate.
• Convection occurs through air movement (cold air entering, hot air leaving).

Conduction

• Conduction is heat transfer between objects in direct contact.
• Example: Heat transferring to your hands when holding a hot cup of coffee.
• The rate of conductive heat transfer is given by: $\frac{Q}{t} = kA\frac{(T<em>2 - T</em>1)}{d}$
  • $\frac{Q}{t}$ = rate of heat transfer (energy per unit time)
  • $d$ = thickness or distance between objects
  • $A$ = contact area
  • $k$ = thermal conductivity
  • $T<em>2 - T</em>1$ = temperature difference

Example: Ice Box

• A styrofoam ice box has a total area of 0.950 m² and walls with an average thickness of 2.50 cm.
• The box contains ice, water, and canned beverages at 0 °C.
• The outside temperature is 35 °C.
• Thermal conductivity of styrofoam: 0.010 J/s·m·°C
• Latent heat of fusion of water: 334 kJ/kg

Convection

• Convection is heat transfer by the macroscopic movement of mass.
• Convection can be natural or forced and is generally faster than conduction.
• Example: Steaming milk for hot cocoa.

Radiation

• Radiation is heat transfer through the emission or absorption of electromagnetic waves.
• Example: Reheating coffee in a microwave.
• The rate of heat transfer by radiation is determined by the Stefan-Boltzmann law: $\frac{Q}{t} = \sigma eAT^4$
  • $\frac{Q}{t}$ = rate of heat transfer (energy per unit time)
  • $\sigma$ = Stefan-Boltzmann constant (5.67 x 10-8 J/s·m²·K⁴)
  • $e$ = emissivity of the body (0 to 1)
  • $A$ = surface area
  • $T$ = temperature in Kelvin
• The net rate of heat transfer by radiation is related to both the object's temperature and its surroundings' temperature:
  • $\frac{Q<em>{net}}{t} = \sigma eA(T</em>1^4 - T_2^4)$

Example: Radiation from a Person

• An unclothed person is standing in a dark room with an ambient temperature of 22.0 °C.
• The person's skin temperature is 33.0 °C, and the surface area is 1.50 m².
• The emissivity of skin is 0.97.
• Stefan-Boltzmann constant: 5.67 x 10-8 J/s·m²·K⁴
• $T_2 = 22°C + 273 = 295 K$
• $T_1 = 33°C + 273 = 306 K$
• $\frac{Q}{t} = (5.67 × 10^{-8} J/s m^2 K^4)(0.97)(1.50 m^2)((306K)^4 - (295K)^4) = - 99 J/s = -99 W$