1.

What will be the final temperature of the mixture as a result of mixing 5g ice at 10 ° C with 22 g water at 30 ° C temperature ? The relative heat of ice is 0.5 cal g-l. ° C-1 .​

Answer»

Explanation:

Given: 5 g of ICE at −10

o

C are mixed with 20g of water at 30

o

C. Specific heat of ice is 0.5 and latent heat of water 80calg

−1

.

To find the final temperature of the mixture

Solution:

We know,

Specific heat of water, s

1

=4.18J/(g

∘

C)=1cal/(g

∘

C)

As per the given condition,

MASS of ice, m

2

=5g

Mass of water, m

1

=20g

Temperature of ice, T

2

=−10

∘

C

Temperature of water, T

1

=30

∘

C

specific heat of ice, s

2

=0.5cal/g.°C

latent heat of water L=80calg

−1

.

Let the final temperature be, T

Here, Heat lost by 20g water = Heat energy needed to change the temperature of ice from –10°C to 0°C + Latent heat needed to change ice at 0°C into water at 0°C + heat absorbed by water (melted ice)

m

1

s

1

(T

1

−T)=m

2

s

2

(0−T

2

)+m

2

L+m

2

s

2

(T−T

2

)

⟹20×1×(30−T)=5×0.5(0−(−10))+5×80+5×1×(T−(0))

⟹600−20T=25+400+5T

⟹25T=600−400−25

⟹T=7

∘

C

is the final temperature

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