Question

a) You wish to melt a 3.0-kg block of aluminum, which has already been heated to 660ºC. How much heat is required to melt the aluminum? (the textbook lists the latent heat of aluminum as 396,000 J/kg)

b) You wish to melt a 3.0-kg block of aluminum, which initially at 20ºC. How much heat is required to melt the aluminum? (the textbook lists the latent heat of aluminum as 396,000 J/kg)

Answer 1

Answer:

(a) Given, mass m = 3.0 kg and Latent heat of aluminium L = 396000 J/kg

The melting point of aluminium is 660 0C. Here the aluminium is already been heated at 660 0C. So the amount of heat required to melt the aluminium is Q = mL

Therefore, Q = (3.0 kg) (396000 J/kg) = 1188000 J or 11.88 x 105 J

(b) Gievn mass m = 3.0 kg, initial temperature Ti = 20 0C = 20 + 273K = 293 K

We need the heat energy to raise its temperature upto its melting point, that means the final temperature is equal to the melting point of aluminium. i.e., Tf = 660 0C = 660 + 273K = 933 K

Specific heat of aluminium is c = 912.096 J/(kg-K)

Therefore, the amount of heat required to its melting point is Q = mcT = mc(Tf - Ti)

Q = (3.0 kg) (912.096 J/(kg-K))(933K - 293K) = 1751224 J

The total heat required to melt the aluminium is mcT + mL

Therefore, 1751224 J + 1188000 J = 2939224 J or 29.39 x 105 J