Question

540g of ice at 0°C is mixed with 540g of water at 80°C. The final temperature of the mixture is..???

Answer 1

`tf=0.086°C`

Explanation:

`mL+mcΔt=-mcΔt`

`(540g)(334J/g)+(540g)*(4.184J/(g°C))(tf-0°)=-(540 g)(4.184 J/(g°C))(tf-80°)`

`180360J+2259.36 J/(°C)*tf-0J= -2259.36 J/(°C)+180748.8J`

`4518.72 J/(°C)*tf=388.8J`

`tf=0.086°C`

Answer 2

`0^@C`

Explanation:

Heat energy liberated by `540 g` of water to cool down to `0^@C` is `540*1*(80-0) C=43200 C` (using, `H = ms d theta` ,where, `m` is the mass, `s` is the specific heat and `d theta` is change in temperature)

Now,heat energy required for `540 g` of ice to get converted to same amount of water at `0^@C` is `540*80=43200 C` (using, `H=ml` where, `l` is the latent heat of melting of ice = `80 C/g`)

So,the required heat energy for the given amount of ice to get converted to water at `0^@C` will be supplied exactly due to the cooling of water to `0^@C`, so total `540+540=1080g` of water will be present at `0^@C`