Question

A thermally insulated pot has 150 g ice at temperature 0 o C. How much steam of 100 o C has to be mixed to it, so that water of temperature 50 o C will be obtained? (Given : latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g 0 C)

Answer 1

Nice Question.

Given Condition ⇒
Mass of the Ice = 150 g.
Latent Heat of the Melting of the Ice = 80 Cal/g


Now, Heat Energy Required to melt 150 g of the Ice = mass of the Ice × Latent Heat.
= m × L
= 150 × 80
= 12,000 Cal.

Now,
Heat Energy Required to convert 150 g of the Water at 0° C to water at 50° C = mass × Specific Heat Capacity × Change in temperature.

∴ Heat Energy = 150 × 1 × (50 - 0)
   = 150 × 50
   = 7500 Cal.

Thus,
Total Heat Energy required to melt the Ice at 0° C to water at 50° C = Heat Energy required to melt Ice + Heat energy required to convert water at 0° C to water 50°C = 12,000 + 7,500
= 19,500 Cal.


Now, In Second Case,

Let the Mass of the steam required to melt the Ice at 0° C and convert it to 50° C be x g.

Heat energy given by the Steam to convert the Ice into water = Mass of the Steam × Latent Heat of Vaporization of the steam.
= x × 540
= 540x

According to the Principle of Calorimetry (or the Principles of the Mixtures), Heat energy released by the steam is equal to the Heat energy gained by the Ice to convert it into 50° C into water.

∴ 19500 = 540x
∴ x = 19500/540
∴ x = 36.11 g.


Hence, the mass of the steam required to provide the heat energy so that the Ice at 0° C into water at 50°C is 36.11 g.



Hope it helps.

Answer 2

Answer:

Here is your answer

Explanation:

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