Question

22320 cal heat is supplied to 100g of ice at 0°C. If the latent heat of fusion of ice is 80cal gl and latent heat of vaporization of water is 540 cal g'', the final amount of water thus obtained and its temperature respectively are

A. 8g, 100°C
B. 100g, 90°C
c. 92g, 100°C
D. 82g, 100°C​

Answer 1

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Answer 2

c. 92g, 100°C

Explanation:

Given:

• amount of heat supplied, $Q=22320\ cal$

• mass of ice initially, $m_i=100\ g$

• initial temperature of ice, $T_{i}=0^{\circ}C$

• latent heat of fusion of ice, $L_f=80\ cal.g^{-1}$

• latent heat of vapourization of  ice, $L_v=540\ cal.g^{-1}$

Now the amount of heat required by 100 gram of ice to convert to  100 gram of water at 0 Celsius.

$q_{f}=m.L_f$

$q_f=100\times 80$

$q_f=8000\ cal$

So, now the remaining amount of heat will raise the temperature of water:

$Q-q_f=m.c.(T_f-T_i)$

$22320-8000=100\times 1\times (T_f-0)$

$T_f=143.2^{\circ}C$ which means that the water will be heated to 100 degree Celsius and the rest of the heat will be used in converting the water to vapor.

So,

$q=100\times 1\times (100-0)$   since the specific heat of water is 1 calorie

$q=10000\ cal$

Now the remaining heat is :

$Q'=Q-q_f-q$

$Q'=22320-8000-10000$

$Q'=4320\ cal$

Now this heat is used to convert the water of 100 degree Celsius into vapour at 100 degree Celsius:

$Q'=m.L_v$

$4320=m\times 540$

$m=8\ g$ is converted into vapour.

Hence the remaining mass of water is (100-8)=92 g at 100 degree Celsius.

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TOPIC: latent heat

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