Temperature and Heat

Overview

Temperature is a measure of the average kinetic energy of particles in a substance. Heat is the transfer of energy between objects at different temperatures.

Temperature Scales

Celsius (°C)

• Water freezes: 0°C
• Water boils: 100°C

Fahrenheit (°F)

• Water freezes: 32°F
• Water boils: 212°F

Kelvin (K)

• Absolute zero: 0 K
• Water freezes: 273.15 K
• Water boils: 373.15 K

Conversions

$$T_K = T_C + 273.15$$ $$T_F = \frac{9}{5}T_C + 32$$ $$T_C = \frac{5}{9}(T_F - 32)$$

Temperature Changes

$$\Delta T_K = \Delta T_C$$ $$\Delta T_F = \frac{9}{5}\Delta T_C$$

Heat

Definition

Heat ($Q$) is energy transferred due to temperature difference.

• Unit: Joule (J)
• Also: calorie (1 cal = 4.186 J)

Sign Convention

• $Q > 0$: heat absorbed by system
• $Q < 0$: heat released by system

Heat Capacity

Heat Capacity ($C$)

Heat needed to raise temperature by 1 K:

$$Q = C\Delta T$$

Specific Heat Capacity ($c$)

Heat capacity per unit mass:

$$Q = mc\Delta T$$

Unit: J/(kg·K) or J/(kg·°C)

Molar Heat Capacity ($C_m$)

Heat capacity per mole:

$$Q = nC_m\Delta T$$

Unit: J/(mol·K)

Specific Heat Values

Substance	$c$ (J/kg·K)
Water	4186
Ice	2090
Steam	2010
Aluminum	900
Copper	387
Iron	449
Lead	128

Water has unusually high specific heat!

Phase Changes

Latent Heat

Energy for phase change at constant temperature:

$$Q = mL$$

Where $L$ = specific latent heat

Latent Heat of Fusion ($L_f$)

Solid ↔ Liquid transition

$$Q = mL_f$$

Latent Heat of Vaporization ($L_v$)

Liquid ↔ Gas transition

$$Q = mL_v$$

Values for Water

• $L_f = 3.34 \times 10^5$ J/kg (ice ↔ water)
• $L_v = 2.26 \times 10^6$ J/kg (water ↔ steam)

Calorimetry

For an isolated system, heat lost = heat gained:

$$Q_{\text{lost}} + Q_{\text{gained}} = 0$$

Or:

$$\sum Q = 0$$

Thermal Equilibrium

When two objects reach the same temperature:

$$m_1 c_1 (T_f - T_1) + m_2 c_2 (T_f - T_2) = 0$$

Solving for final temperature:

$$T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2}$$

Examples

Example 1: Heating Water

Heat 2 kg of water from 20°C to 80°C.

$$Q = mc\Delta T = 2 \times 4186 \times 60 = 502{,}320 \text{ J} \approx 502 \text{ kJ}$$

Example 2: Melting Ice

Melt 0.5 kg of ice at 0°C.

$$Q = mL_f = 0.5 \times 3.34 \times 10^5 = 1.67 \times 10^5 \text{ J}$$

Example 3: Complete Phase Change

Convert 1 kg ice at −10°C to steam at 110°C.

1. Heat ice: $Q_1 = mc_{\text{ice}} \times 10 = 1 \times 2090 \times 10 = 20{,}900$ J
2. Melt ice: $Q_2 = mL_f = 1 \times 334{,}000 = 334{,}000$ J
3. Heat water: $Q_3 = mc_{\text{water}} \times 100 = 1 \times 4186 \times 100 = 418{,}600$ J
4. Vaporize: $Q_4 = mL_v = 1 \times 2{,}260{,}000 = 2{,}260{,}000$ J
5. Heat steam: $Q_5 = mc_{\text{steam}} \times 10 = 1 \times 2010 \times 10 = 20{,}100$ J

Total: $Q = 3{,}053{,}600$ J $\approx 3.05$ MJ

Example 4: Calorimetry

100 g of copper at 200°C is placed in 200 g of water at 20°C. Find equilibrium temperature.

$$m_{Cu} c_{Cu} (T_f - 200) + m_w c_w (T_f - 20) = 0$$ $$0.1 \times 387 \times (T_f - 200) + 0.2 \times 4186 \times (T_f - 20) = 0$$ $$38.7 T_f - 7740 + 837.2 T_f - 16744 = 0$$ $$875.9 T_f = 24484$$ $$T_f = 28°\text{C}$$

Example 5: Ice in Water

50 g of ice at 0°C is added to 200 g of water at 25°C. Find final state.

Heat available from water cooling to 0°C:

$$Q_{\text{available}} = 0.2 \times 4186 \times 25 = 20{,}930 \text{ J}$$

Heat needed to melt all ice:

$$Q_{\text{needed}} = 0.05 \times 334{,}000 = 16{,}700 \text{ J}$$

Since $Q_{\text{available}} > Q_{\text{needed}}$, all ice melts.

Remaining heat warms the mixture:

$$(0.2 + 0.05) \times 4186 \times T_f = 20{,}930 - 16{,}700$$ $$T_f = \frac{4{,}230}{0.25 \times 4186} = 4°\text{C}$$