# How much heat in joules must be added to 0.841 kg of aluminu

• Alice7979
In summary, the question is asking for the amount of heat in joules needed to change 0.841 kg of aluminum from a solid state at 130 °C to a liquid state at 660 °C, with a melting point of 660 °C. The equation used is Q = mL, and the total heat needed is calculated using the specific heat of aluminum (0.9 J/g) and the heat of fusion (400,000 J/kg). After correcting a mistake in the original attempt at a solution, the final answer is not yet determined.

## Homework Statement

How much heat in joules must be added to 0.841 kg of aluminum to change it from a solid at 130 °C to a liquid at 660 °C (its melting point)? The latent heat of fusion for aluminum is 4.0 x 105 J/kg.

## Homework Equations

Q = mL
Total = mc∆t + mL = m(c∆t +l)

## The Attempt at a Solution

i think it is it .841(.9(660-130) + 4*105) but the answer i get isn't right

Alice7979 said:

## The Attempt at a Solution

i think it is it .841(.9(660-130) + 4*105) but the answer i get isn't right
That 0.9, the specific heat of aluminum, is not in standard units. It’s 0.9 kilojoules / kg (or equivalently, 0.9 J / g) and not 0.9 J / kg

Alice7979
Also, that heat of fusion should be 400000 J/kg (actually 321000 J/kg according to an internet source).

Nathanael
Chestermiller said:
Also, that heat of fusion should be 400000 J/kg (actually 321000 J/kg according to an internet source).
I thought it was weird to say 4*105 hahaha 10^5 makes a lot more sense

Nathanael said:
That 0.9, the specific heat of aluminum, is not in standard units. It’s 0.9 kilojoules / kg (or equivalently, 0.9 J / g) and not 0.9 J / kg
Nathanael said:
I thought it was weird to say 4*105 hahaha 10^5 makes a lot more sense
Yea I didn't realize it changed it to that.