10 grams of ice at - 20 degree celsius is dropped into a calorimeter containing 10 grams of water at 10 degree celsius. The specific heat of water is twice that of ice. when equilibrium is reached, the calorimeter will contain?
Answers
We have ice at -10°C,
For the ice to come to 0°C, the energy required is m*s*∆T
m is mass of ice = 10g
s is specific heat of ice = 0.8 cal/g.°C
∆T is temperature difference = 0-(-10) = 10°C
So the energy required is 10*0.8*10 = 80 cal
The energy released when 50 grams of water at 15°C becomes water at 0°C is m*s*∆T
s for water is 1 cal/g.°C
Energy released is 50*1*(15–0) = 50*15 = 750 cal
For the ice to completely turn into water at 0°C, the energy required is m*L
L is latent heat of ice = 80 cal/g
Energy required is 10*80 = 800 cal.
But the water when turns into water at 0°C, the energy released is only 750 cal.
So, the total ice will not get converted into water.
The energy left after the ice gets to 0°C is 750–80=680 cal
With 680 cal, the amount of ice that turns into water is 680/80 = 8.5 grams.
So the composition is 10–8.5=1.5 grams of ice at 0°C and 10+8.5=18.5 grams of water at 0°C.
- By Subhadeep Maji
Answer:
Correct answer is : 10 gm ice and 10 gm of water
Explanation:
See the attachment