How much heat must be added to 0.55 kg of aluminum to change it from a solid at 180°C to a liquid at 660°C (its melting point)? The latent heat of fusion for aluminum is 4.0x10^5 J/kg.

I understand the Equation Q=mc times change it T
would that be .55 times 4.0 x 10^5 times (660-180) or im i mising something please help im still haveing some trouble with this problem

FIRST you must heat the Al from 180 to 660

for that you need the heat capacity of Al, call it Cal, which is in your book.
Q1 = .55 * Cal * (660-180)
THEN you melt it
Q2 = .55 * 4*10^5
period, no temp in that eqn it happens at melting point

Then add Q1+Q2

To calculate the amount of heat required to change a substance from a solid to a liquid, you need to account for two types of energy transfers: the heat required to raise the temperature of the solid from 180°C to its melting point (660°C), and the heat required for the phase change from solid to liquid.

Let's break down the calculation step by step:

Step 1: Calculate the heat required to raise the temperature of the solid aluminum:
Q1 = mcΔT1
Where:
m = mass of aluminum = 0.55 kg
c = specific heat capacity of aluminum = 900 J/kg·°C (approximately)
ΔT1 = temperature change = final temperature (660°C) - initial temperature (180°C)

Q1 = 0.55 kg * 900 J/kg·°C * (660°C - 180°C)
= 0.55 kg * 900 J/kg·°C * 480°C

Step 2: Calculate the heat required for the phase change from solid to liquid:
Q2 = mLf
Where:
m = mass of aluminum = 0.55 kg
Lf = latent heat of fusion for aluminum = 4.0 x 10^5 J/kg

Q2 = 0.55 kg * 4.0 x 10^5 J/kg

Step 3: Calculate the total heat required:
Q_total = Q1 + Q2

So the total heat required to change 0.55 kg of aluminum from a solid at 180°C to a liquid at 660°C is given by:

Q_total = (0.55 kg * 900 J/kg·°C * 480°C) + (0.55 kg * 4.0 x 10^5 J/kg)

Now you can compute the value of Q_total.