Solid aluminum has a specific heat of 0.89 J/g oC, and a heat of fusion of 398 J/g. Aluminum melts at 660.4oC. How much heat is required to convert a 10.0 g sample of aluminum metal at 580.0oC to a fully-molten sample of liquid at 660.4oC?

Solid aluminum has a specific heat of 0.89 J/g oC, and a heat of fusion
of 398 J/g. Aluminum melts at 660.4oC. How much heat is required to convert a
10.0 g sample of aluminum metal at 580.0oC to a fully-molten sample of liquid at
660.4oC?

Answer 1

I found: #4695.6J#

Here you have a bloc of solid aluminium and you heat it up to reach the temperature of fusion and then you supply a bit more of heat to completely melt it. So in total you have: #Q_1+Q_2=# Where:

#Q_1=mc_"Al"DeltaT# is the heat needed to heat up your bloc from #580^@C# to #660.4^@C#

#Q_2=mL_f# is the heat needed not to increase its temperature but to melt it completely.

We possess:

#Q_1+Q_2=mc_"Al"DeltaT+mL_f=#

#=10*0.89*(660.4-580)+10*398=4695.6J#

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Answer 2

Henry Antrim

To calculate the total heat required, we need to consider two steps: heating the solid aluminum to its melting point and then melting it.

Heat required to raise the temperature of the solid aluminum: ( Q = m \times c \times \Delta T ) ( Q = 10.0 , \text{g} \times 0.89 , \text{J/g}^\circ \text{C} \times (660.4 - 580.0)^\circ \text{C} )
Heat required to melt the aluminum: ( Q = m \times \text{heat of fusion} ) ( Q = 10.0 , \text{g} \times 398 , \text{J/g} )

Total heat required: ( Q_{\text{total}} = Q_1 + Q_2 )

Calculate ( Q_1 ) and ( Q_2 ), then add them together to get ( Q_{\text{total}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.