Question

One end of a steel rod (K = 46 J s⁻¹ m⁻¹°C⁻¹) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm². Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹.

Answer 1

Given:
Thermal conductivity of the rod, K = 46 J s^–1 m^–1 °C^–1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A = 0.04 cm²
= 0.04 × 10⁻⁴ m²
= 4 × 10⁻⁶ m²
Rate of transfer of heat is given by
$$\begin{gathered} \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\Delta T}{{\displaystyle \frac{l}{\mathrm{kA}}}} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\frac{\left(T_1-T_2\right)kA}{l} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=\left(\frac{100-0}{1}\right)\times 46\times 4\times {10}^{–6} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t}=184\times {10}^{-4}\mathrm{J}/\mathrm{s} \end{gathered}$$


$$\begin{gathered} \mathrm{Also},\mathrm{\Delta Q}=m{\mathrm{L}}_f \\ \therefore \left(\frac{m}{t}\right)L_f=184\times {10}^{-4} \\ \Rightarrow \left(\frac{m}{t}\right)\times 3.36\times {10}^5=184\times {10}^{–4} \\ \Rightarrow m=\frac{184\times {10}^{-4}}{3.36\times {10}^5}\times t \\ \Rightarrow m=5.5\times 10\times {10}^{-9}\times 1\mathrm{kg}/\mathrm{s} \\ \Rightarrow m=5.5\times {10}^{-8}\times {10}^3\mathrm{g}/\mathrm{s} \\ \Rightarrow m=5.5\times {10}^{-5}\mathrm{g}/\mathrm{s} \end{gathered}$$