Question

how much energy required to convert 5gr of ice at-20degree c to 5gr of water vapour of 100 degree c? ​

Answer 1

Answer:

5 g of water at 30

o

C and 5g of ice at −20

o

C are mixed in a calorimeter, water equivalent of calorimeter is negligible. Specific heat of ice is 0.5cal/(g

∘

C) and latent heat of ice is 80 cal per gram.

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

=5g

Temperature of ice, T

2

=−20

∘

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 5g water = Heat energy needed to change the temperature of ice from –20°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

)

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

⟹150−5T=50+400+5T

⟹10T=150−50−400

⟹T=−30

∘

C

So the final temperature of the mixture is 30

∘

C