Question

A metal rod 0.314 m long and 0.02 m in diameter has one end in boiling water at 373K and the other end in melting ice at 273K. Find the amount of ice that will melt per hour? K = 385 Wm^-1K^-1. Latent heat of Ice = 3.36 × 10⁵ Jkg^-1.

Answer 1

If ‘m’ is mass of ice that melts in one hour, then quantity of heat conducted is Q = mL

Here, K = 385Wm^-1K^-1

A = p r² =3.14 × (0.01)²

L = 3.3 × 10⁵ Jkg^-1

θ₁ – θ₂ = 373 – 273 = 100 K

t = 1 hour = 3600 s

d = 0.314 m

= 41.28 x 10^-2kg