Question

Calculate the thermal resistivity of a copper rod 20.0 cm. length and 4.0 cm.
in diamter.
Thermal conductivity of copper = 9.2 x 10–2 temperature different acrosss
the ends of the rod be 50°C. Calculate the rate of heat flow.

Answer 1

length of a copper, ∆x = 20cm

cross sectional area of copper wire, A = πr² = 3.14 × (2cm)² = 3.14 × 4 cm²

= 12.56 cm²

thermal conductivity of copper, k = 9.2 × 10^-2 cal/cm/sec/°C

temperature difference, ∆T = 50°C

using formula,

thermal resistivity, R = ∆x/KA

= 20cm/(9.2 × 10^-2cal/cm/sec/°C × 12.56 cm²)

= 20 × 100/(9.2 × 12.56 cal/sec/°C)

= 17.3 sec.°C/cal

heat flow through copper wire, dQ/dt = KA ∆T/∆x

= ∆T/{∆x/KA}

= 50°C/{17.3 sec.°C/cal}

≈ 3 cal/sec