The specific heat of fusion of lead is 22.6 kJ / Kg. In order to melt 0.6 kg

The specific heat of fusion of lead is 22.6 kJ / Kg. In order to melt 0.6 kg of lead taken at the melting temperature in 10 minutes, the power of the heater must be ….

Initial data: t (duration of lead melting) = 10 min (600 s); m (lead mass) = 0.6 kg; the lead was at its melting point.

Reference values: according to the condition λ (specific heat of fusion of lead) = 22.6 kJ / kg (22.6 * 10 ^ 3 J / kg).

The power of the heater can be calculated using the formula: N = A / t = Q / t = λ * m / t.

Let’s make the calculation: N = 22.6 * 10 ^ 3 * 0.6 / 600 = 22.6 watts.

Answer: To melt lead, the heater power must be 22.6 W.