Consider a 0.63-kg sample of metal at room temperature of 20
c. the addition of 642,000 j increases its temperature to its melting point (782
c. an additional 59,400 j causes the sample to completely liquefy. what is the heat of fusion of the sample?

It is to be noted that to answer this question, we only use the given mass and the given amount of heat needed to liquefy the object. The equation that we can use is,
                               Hf = mhf
where Hf is the amount of heat, m is the mass, and hf is the heat of fusion. 
                               59400 J = (0.63 kg)hf
The value of hf from the equation is 94,285.71 J/kg.

Sorry my friend i don't know the answer. i am just in 6th