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?

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.

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

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

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

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.

To calculate the amount of energy required to change 150 grams of substance A from a solid at -40 degrees Celsius to a liquid at 70 degrees Celsius, we need to consider the different energy changes involved.

Step 1: Calculate the energy required to heat the solid from -40 degrees Celsius to its fusion point (-10 degrees Celsius).
The temperature change is ΔT = -10 degrees Celsius - (-40) degrees Celsius = 30 degrees Celsius.
The energy required to heat the solid is given by:
Q = mass × specific heat of the solid × temperature change
= 150 g × 3 J/g°C × 30°C
= 13,500 J or 13.5 kJ (since 1 kJ = 1000 J)

Step 2: Calculate the energy required to change the solid at its fusion point to a liquid at the same temperature.
The energy change during fusion is given by:
Q = mass × enthalpy of fusion
= 150 g × 150 J/g
= 22,500 J or 22.5 kJ

Step 3: Calculate the energy required to heat the liquid from its fusion point to the final temperature of 70 degrees Celsius.
The temperature change is ΔT = 70 degrees Celsius - (-10) degrees Celsius = 80 degrees Celsius.
The energy required to heat the liquid is given by:
Q = mass × specific heat of the liquid × temperature change
= 150 g × 6.2 J/g°C × 80°C
= 74,400 J or 74.4 kJ

Step 4: Sum up the energies from all three steps to determine the total energy required:
Total energy = Energy to heat the solid + Energy during fusion + Energy to heat the liquid
= 13.5 kJ + 22.5 kJ + 74.4 kJ
= 110.4 kJ

Therefore, it will require approximately 110.4 kJ of energy to change 150 grams of substance A from a solid at -40 degrees Celsius to a liquid at 70 degrees Celsius.