Question

20g of solid at 20c is mixed with 100 g pf cold water at 5c if the finql tempt of mixture is 10c find specific heat capacity of solid?
​

Answer 1

Question

What would be the final temperature of the mixture when 5g 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 80cal g

−1

.

Medium

Solution

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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.

Answer 2

mct=mct

20g x c x (20-10) = 100g x 4.2j/gc x (10-5)

20 x c x 10 = 10 x 42 x 5

c = 420 x 5/200

c = 21/2

c = 10.5j/gc