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Heat Transfer

In an insulat...

Question

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, $L_{f u s i o n} = 80 c a l / g = 336 J / g,$
$L_{v a p o r i z a t i o n} = 540 c a l / g = 2268 J / g,$
$S_{i c e} = 2100 J / k g K = 0.5 c a l / g K$ and
$S_{w a t e r} = 4200 J / k g K = 1 c a l / g K$

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Solution

$Σ △ θ = 0$
Heat lost by steam to convert into water at $0^{o} C$
$H_{L} = 0.05 \times 540 + 0.05 \times 100 \times 1 = 27 + 5 = 32 K c a l$
Heat required to change ice into $0^{o} C$ water.
$H_{g} = 0.45 \times 0.5 \times 20 + 0.45 \times 80 = 4.5 + 36 = 40.5 K c a l$

Released heat is less than the required heat so final temperature is $0^{o} C$ of the mixture.

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Similar questions

0.05 k g

373 K

0.45 k g

253 K

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

100^{o} C

(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)

1 g

100^{\circ} C

0^{\circ} C

= 540 c a l / g

= 80 c a l / g

= 1 c a l / g^{\circ} C

1 kg

0^{\circ} C

1 kg

100^{\circ} C

[L_{f u s i o n, i c e} = 3.36 \times 10^{5} J/kg,

L_{v a p o r i z a t i o n, w a t e r} = 2.26 \times 10^{6} J/kg,

C_{w a t e r} = 4200 J/kg-K]

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