Question

Does anyone know how to work this Celsius equation?

How much heat is necessary to change 455 g of ice at -5°C to water at 20°C?

Answer 1

`46637.5` Calorie

Explanation:

In this process first we need to convert the ice at `-5 ^@C` to `0^@C`,so heat energy required is `455*0.5*(0-(-5)) J=1137.5` Calorie (using `H=ms d theta`,where, `m` is the mass of ice with specific heat `s`(for ice it's `0.5` CGS units) and that has undergone a change in temperature of `d theta`)

Next,we need to supply the required latent heat to change its state from ice to equal amount of water at `0^@ C`,

Now,latent heat required for melting of ice = `80 C/g`

So,for `455g` latent heat required will be `455*80=36400` Calorie.

Now,again we have to supply heat energy to convert water at `0^@C` to `20^@C`,so for that process,heat energy required will be `455*1*(20-0)=9100` Calorie (using `H=ms d theta`)

So,heat energy required for the entire process is the sum of the above `3` values i.e `1137.5+36400+9100=46637.5` Calorie

Answer 2

I got almost `200kJ`

Explanation:

Have a look: