Question

100g of ice at 0^oC and 100g of steam at 100^oC are placed in a well-insulated container at atmospheric pressure.

a) How much heat would be required to melt all of the ice and raise the resulting water temperature to 100^oC?

b) How much heat would need to b removed from steam to condense all of it to water?

c) Based on the answers in previous parts, when equilibirum is reached, will the final state be

   i) mixture of ice and water

    ii) mixture of ice and steam

   iii) all water. ?

d) Determine the final equilibrium temperature of the system, and determine each of the masses of ice, water, and/or steam in the final state.

Answer 1

Part (a)

To melt all the ice and convert it into 100° C water we have to go through the process as:

100g, 0°C ice- - >100g, 0°C water

100g, 0°C water - - >100g, 100°C water

So required heat is

Q = mL_f + mcdT

Q = 100×80 + 100 ×1×100

Q =18000 calories

Part (b)

Heat required to condensed all the steam into water is

Q = mL_v​​​​​​

Q = 100×540

Q = 54000 calories

Part (c)

As from above results we can say that to convert 0°C ice into 100°C water whole steam will not be condensed because energy released is more than energy absorbed. So, when equilibrium reach final state will contain mixture of (water + steam)

Part (d)

As the final mixture is of water at 100°C + steam at 100°C

So final temperature of mixture is 100°C

So mass of ice = 0g

To covert 0°C ice into 100°C water we required 18000 calories so mass of steam condensed will be :

mL_v = 18000 calories

m×540 = 18000

m = 33.33 g

Therefore,

Mass of water (m_w​​​​​​) = 100 +(condensed steam)

m_w = 100 + 33.33 = 133.33 g

Mass of steam (m_s​​​​​​) = 100 - (condensed steam)

m_s = 100 - 33.33 = 66.67 g