Question

Calucalation in change in coton
when 5kg. water at 3öc is
mixed with 15gm water at 10°C​

Answer 1

Explanation:

Answer

Given: 2 kg of ice at −20

∘

C is mixed with 5 kg of water at 20

∘

C. It is given that the specific heats of water and ice are 1cal/(g

∘

C) and 0.5cal/(g

∘

C) respectively and the latent heat of fusion of ice is 80cal/g.

To find the final mass of water in the mixture.

Solution:

As per the given criteria,

Mass of water, m

w

=5kg

Mass of the ice, m

i

=2kg

Temperature of ice, T

i

=−20

∘

C

Temperature of water, T

w

=20

∘

C

Specific heat of ice, s

i

=0.5cal/(g

∘

C)

Specific heat of water, s

w

=1cal/(g

∘

C)

latent heat of fusion of ice is L=80cal/g.

Let M be the final mass of the content

The amount of heat losed by water at 0 degreeis

Q

w

=m

w

s

s

(ΔT)=5×1×20=100kcal

the amount of heat gained by ice to go to 0 degree is

Q

i

=m

i

s

i

(ΔT)=2×0.5×20=20kcal

heat left for absorption is

Q

w

−Q

i

=100−20=80kcal

This is equal to latent heat, i.e.,

mL=80

⟹m=

L

80

=

80

80

=1kg

This is the mass of water from the converted ice.

So the final mass of water in the vessel is , M=m

w

+m=5+1=6kg