How much heat is required to melt 20 g of lead at a temperature of 27 degrees?

Initially, lead needs to be heated from 27 ° C to 327 ° C – to its melting point. And then melt it. The total amount of required heat will consist of two parts – heat of heating and heat of fusion – Q = Q load + Q melt.

The amount of heat that needs to be reported to lead will be determined by the formula – Q load = c * m * (t load – t start), where Q load is the amount of heat required to heat the body, c is the specific heat capacity of the substance (in our case, lead) , m – body weight, t load –
the temperature to which the body was heated, t start – the initial temperature at which the bodies began to be heated.

(t heat – t start) – the temperature difference in our case is 327 ° С – 27 ° С = 300 ° С.

The specific heat capacity of lead is 140 J / kg ° C.

20 g of lead means a mass of 0.02 kg.

Substituting the values ​​into the above formula, we get – Q load = (140 J / kg ° C) * (0.02 kg) * (300 ° C) = 840 J.

In order to melt lead at the melting temperature, it will be necessary to report the amount of heat, which is determined by the formula – Q melt = λm, where Q melt is the amount of heat required to melt lead, λ is the specific heat of melting of lead (25 kJ / kg), m is the mass – Q melt = 25 kJ / kg * 0.02 kg = 0.5 kJ or 500 J.

The final answer: the total amount of thermal energy required to melt ice under the given conditions will be – Q = Q load + Q melt – 840 kJ + 500 J = 1340 J.