Question

The specific heat of ice is 0.492 cal/(g × °C). How many calories of heat are required to raise 100.0 g of ice from -20.0 °C to -0.5 °C?

Answer 1

Well, your ice DOES NOT undergo a phase change....

Explanation:

And so we simply take the product...

`"mass"xx"C"_"specific heat"xxDeltaT=100.0*gxxunderbrace(0.492*cal*g^-1*""^@C^-1)_"quoted specific heat"xx19.5*""^@C=+959.4*cal`....the positive sign indicates heat is ADDED to the system.

Answer 2

`959.4` calories

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 substance in kilograms

• `c` is the specific heat capacity of the object in joules

• `DeltaT` is the change in temperature

Here, `DeltaT=-0.5^@"C"-(-20^@"C")=19.5^@"C"`.

So, we get:

`q=100color(red)cancelcolor(black)"g"*(0.492 \ "cal")/(color(red)cancelcolor(black)"g"color(red)cancelcolor(black)(""^@"C"))*19.5^@"C"`

`=959.4 \ "cal"`