Substance A has a normal fusion point of -10 degrees C, an enthalpy of fusion = 150 Jg^-1; specific heats for the solid and the liquid are 3 and 6.2 Jg^-1C^-1 respectively. To change 150 grams of A from a solid at -40 degrees C to a liquid at 70 degrees C will require how many kJ?

Answer 1

q1=heat to change T of solid from -40 to -10.
q1 = mass x specific heat solid x (Tf-Ti) where Tf is final T (-10) and Ti is initial T = -40 C.

q2=heat to melt the solid at -10 C to a liquid.
q2= mass x heat fusion.

q3 = heat to change T from -10 to 70.
q3 = mass x specific heat liquid x (Tf-
Ti) =

Q = q1 + q2 + q3.

Answer 2

I don't know how that screwball title got in but the post of how to work the problem is correct.

Answer 3

I see I have a new name, also. Hope this corrects it.

Answer 4

I followed through all of your steps but I am not getting any of the multiple choice answers. The answers that are listed are 91.8, 100.8, 105, 110.4, 119.4 kJ. When I worked the problem out following your steps I came up with the answer 138.3

Answer 5

I worked it out and obtained one of your answers listed. I'll be you are making an algebra error in q1. That is
150 x 3 x [-10-(-40)] =
150 x 3 x +30 = 13,500.
By the way, if you didn't type all of the zeros after 3 and 6.2 and 10 etc etc, then the answer can't come out to four significant figures.