Karena kondisi steady-state, 1D, tanpa pembangkitan panas internal, dan properti material konstan, profil suhu berbentuk garis lurus (linier):
$$\frac{dT}{dx} = \frac{T_2 – T_1}{L}$$
$$\frac{dT}{dx} = \text{konstan}$$
Data Pertama ($T_1 = 50^\circ \text{C}, T_2 = -20^\circ \text{C}$)
Kita hitung berapa laju suhu terhadap posisi
$$\frac{dT}{dx} = \frac{T_2 – T_1}{L}$$
$$\frac{dT}{dx} = \frac{-20 – 50}{0,35}$$
$$\frac{dT}{dx} = \frac{-70}{0,35}$$
$$\frac{dT}{dx} = 200 \text{ K/m}$$
Data Kedua ($T_1 = -30^\circ \text{C}, T_2 = -10^\circ \text{C}$)
Kita hitung berapa laju suhu terhadap posisi
$$\frac{dT}{dx} = \frac{T_2 – T_1}{L}$$
$$\frac{dT}{dx} = \frac{-10 – (-30)}{0,35}$$
$$\frac{dT}{dx} = \frac{20}{0,35}$$
$$\frac{dT}{dx} = 50,142 \text{ K/m}$$
Data Ketiga ($T_1 = 70^\circ \text{C}, \frac{dT}{dx} = 160 \text{ K/m}$)
Kita hitung berapa suhu kedua
$$\frac{dT}{dx} = \frac{T_2 – T_1}{L}$$
$$160 = \frac{T_2 – 70}{0,35}$$
$$56 = T_2 – 70$$
$$T_2 = 126^\circ \text{C}$$
Data Keempat ($T_2 = 40^\circ \text{C}, \frac{dT}{dx} = -80 \text{ K/m}$)
Kita hitung berapa suhu pertama
$$\frac{dT}{dx} = \frac{T_2 – T_1}{L}$$
$$-80 = \frac{40 – T_1}{0,35}$$
$$-28 = 40 – T_1$$
$$T_1 = 68^\circ \text{C}$$
Data Kelima ($T_2 = 30^\circ \text{C}, \frac{dT}{dx} = 200 \text{ K/m}$)
Kita hitung berapa suhu pertama
$$\frac{dT}{dx} = \frac{T_2 – T_1}{L}$$
$$200 = \frac{30 – T_1}{0,35}$$
$$70 = 30 – T_1$$
$$T_1 = -40^\circ \text{C}$$
