Question

200g of ice at -20 degrees celsius is mixed with 500g of water at 20 degree celsius in an insulating vessel

Answer 1

final temperature of mixture = 0°C

mass of water in final mixture = 600g

mass of ice in final mixture = 100g

200g of ice at -20°C is mixed with 500g of water at 20°C in an insulating vessel.

then we have to find final temperature.

we know, heat flows from higher temperature to lower temperature.

so here heat lost by water must be equal heat gained by ice.

at 20°C to 0°C , water lost the heat , h = ms∆T

= 500g × 1cal/g.°C × (20°-0°C)

= 500 × 20

= 10000 cal

and at -20°C to 0°C, heat gained by ice h' = m's'∆T'

= 200g × 0.5cal/g.°C × (0°-(-20°C))

= 100 × 20

= 2000 cal

and heat gained by ice to melt , h" = m'Lf

= 200g × 80cal/g

= 16000 cal

here we see, heat lost by water , h = 10000 cal

and heat gained by ice to melt completely = h' + h" = 2000 + 16000 = 18000 cal

i.e., h > h' + h'

so it is clear that ice doesn't melt completely. hence, final temperature of the mixture must be 0°C.

let x g of ice is melted.

now, heat lost = heat gained

⇒10000 cal = 2000 cal + x g × 80 cal/g

⇒8000 cal = 80x

⇒x = 100g

so, mass of water in the mixture = (500 + 100) = 600g

and mass of ice in the mixture = (200 - 100) = 100g

also read similar questions: 50g of ice at 0 degree Celsius is mixed with 40g of water at 60 degree celsius then ratio of ice and water at equilibriu...

https://brainly.in/question/3280450

If water of temperature 80 degree Celsius and 60 degree Celsius is cooled in two different vessels, which water gets coo...

https://brainly.in/question/5535083