Question

10 gm. of ice at 0°c is mixed with 10 gm. of water at 10°c. temperature of the mixture will be

Answer 1

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.
pls brainlest mark

Answer 2

Answer:

• As per this question, 10 grams of ice at -10°C is mixed with 10 grams of water at 0°C. So, net mass of the mixture = (10 + 10) grams = 20 grams.

• Since, the two constituents of the mixture are at different temperatures; so, there will be exchange of heat between them before attaining equilibrium.

• It is assumed that the mixture is being formed under atmosphere pressure. So, the freezing point of water (or melting point of ice) is considered as 0°C. So, we need to take the amount of latent heat as well; since change of state is also on the card. We know, latent heat of melting of ice is 80 Calories / gram. Specific heats of both ice and water are same (equal to 1).