Question

B) Calculate the energy required to heat up 50 kg of aluminium from 200C to 3000C. The specific heat capacity is 913 J/kg K. ?

Answer 1

I make it approximately `140.29` megajoules.

Explanation:

We use the specific heat equation, which states that,

`q=mcDeltaT`

where:

• `q` is the heat energy supplied in joules

• `m` is the mass of the object in kilograms

• `c` is the specific heat capacity of the object, here it's in joules per kilogram kelvin

• `DeltaT` is the change in temperature

Here, `DeltaT=3000^@"C"-200^@"C"=2800^@"C"=3073.15 \ "K"`.

So, we get:

`q=50color(red)cancelcolor(black)"kg"*(913 \ "J")/(color(red)cancelcolor(black)"kg"color(red)cancelcolor(black)"K")*3073.15color(red)cancelcolor(black)"K"`

`=140289298 \ "J"`

`~~140.29 \ "MJ"`

Now that's a lot of heat energy!