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Principle of Calorimetry

The final tem...

Question

The final temperature, when 10 g of steam at $100^{o} C$ is passed into an ice block of mass 100 g
$(L_{s t e a m} = 540 c a l / g, L_{i c e} = 80 c a l / g; S_{w a t e r} = 1 c a l / g^{o} C)$ is

A

0^{0} C

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B

{15.7}^{o} C

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C

{16.9}^{o} C

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D

{20.4}^{o} C

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Solution

The correct option is A $0^{0} C$
As the man of steam is for small compared to ice. Lets assume that final temp be $0^{o} C$ .
If
$Δ H$ released $< Δ H$ melting
Assumption is right else wrong
So,
$Δ H$ melting $= m_{i c e} \times L_{i c e}$
$Δ H$ released $= m_{s t e a m} \times L_{s t e a m}$
$+ (m_{s t e a m}) (S_{w a t e r}) | Δ T |$
$= (10) (540) + 10 \times 1 \times 100$
$= 5500 c a l$
As $Δ H$ released $Δ H$ melting, assumption is right.

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Similar questions

Q. In an insulated vessel,

0.05 k g

373 K

0.45 k g

253 K

are mixed. Then, find the final temperature of the mixture.
Given,

L_{f u s i o n} = 80 c a l / g = 336 J / g

L_{v a p o r i s a t i o n} = 540 c a l / g = 2268 J / g

S_{i c e} = 2100 J / k g K = 0.5 c a l / g K

S_{w a t e r} = 4200 J / k g K = 1 c a l / g K

5 g

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6 g

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. If the latent heat of steam and ice are

540 cal/g

80 cal/g

respectively and specific heat capacity of water is

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Q. In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Then, find the final temperature of the mixture.
Given,

L_{f u s i o n} = 80 c a l / g = 336 J / g,

L_{v a p o r i z a t i o n} = 540 c a l / g = 2268 J / g,

S_{i c e} = 2100 J / k g K = 0.5 c a l / g K

S_{w a t e r} = 4200 J / k g K = 1 c a l / g K

1 g

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= 540 c a l / g

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100^{o} C

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[ Heat of fusion and heat of vaporization for

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