Temperature Scales and Heat Transfer
Temperature and heat transfer form the gateway to thermodynamics. Temperature scales — Celsius, Fahrenheit, and Kelvin — are related by simple linear conversions, while heat transfer between objects follows the calorimetry equation Q = mcΔT. Understanding these relationships is essential before tackling gas laws, engine cycles, and entropy. Whether you are sizing a home heating system or analyzing a chemical process, every calculation begins with tracking how thermal energy moves from hot to cold.
Temperature Scales and Heat Transfer
Temperature conversions are linear functions, and heat transfer uses simple multiplication.
$$T_F = \frac{9}{5}T_C + 32, \qquad T_C = \frac{5}{9}(T_F - 32)$$
$$T_K = T_C + 273.15$$
$$Q = mc\Delta T$$
$Q$ = heat energy (J), $m$ = mass (kg), $c$ = specific heat (J/(kg·K)), $\Delta T$ = temperature change.
Convert 100°C to Fahrenheit and Kelvin.
$T_F = (9/5)(100) + 32 = 212$°F.
$T_K = 100 + 273.15 = 373.15$ K.
How much heat is needed to raise 2 kg of water from 20°C to 80°C? ($c_{\text{water}} = 4{,}186$ J/(kg·K))
$Q = mc\Delta T = 2(4186)(60) = 502{,}320$ J ≈ 502 kJ.
50 g of iron (c = 450 J/(kg·K)) at 200°C is dropped into 200 g of water at 20°C. Find equilibrium temperature.
Heat lost = heat gained: $m_{Fe}c_{Fe}(200 - T) = m_w c_w (T - 20)$
$0.05(450)(200 - T) = 0.2(4186)(T - 20)$
$22.5(200 - T) = 837.2(T - 20)$
$4500 - 22.5T = 837.2T - 16744$
$21244 = 859.7T$ → $T \approx 24.7$°C.
Practice Problems
Show Answer Key
1. $(9/5)(37) + 32 = 98.6$°F
2. $(5/9)(32 - 32) = 0$°C
3. $-273.15$°C
4. $Q = 5(4186)(10) = 209{,}300$ J
5. $\Delta T = 9000/(2 \times 450) = 10$°C
6. $T = (9/5)T + 32$ → $-4T/5 = 32$ → $T = -40$°