How much heat is required to melt 300 g of tin taken at a temperature of 64 ° C?

Q = Q1 + Q2.

Q1 (heating of gold) = С * m * (tmelt – tн); С – specific heat capacity (tab. Value С = 130 J / (kg * K)), m – mass (by condition m = 300 g = 0.3 kg), tmelt – melting point (tab.value tm = 1064 ºС) , tн – initial temperature (by condition tн = 64 ºС).

Q2 (gold melting) = λ * m; λ – specific heat of fusion (tab.value λ = 0.67 * 10 ^ 5 J / kg).

Q = 130 * 0.3 * (1064 – 64) + 0.67 * 10 ^ 5 * 0.3 = 59100 J = 59.1 kJ.

Answer: To melt 300 g of gold, you need 59.1 kJ of heat.