Question

1.700x1081.700x108 J of energy is removed from 60.0 kg of steam initially at 260°F. You may assume that the steam is cooled at constant pressure and that the specific heat of the steam is 1850 J/kg°C.

A)What is the final temperature? T=? C

What is the final phase? Answer: Combintation of Liquid Water/Ice

B) What mass of water has been converted into ice? m=? kg

Answer 1

Given : Energy removed =

m = mass of steam = 60 kg

T1 = initial temperature = 260 = 126.67

s = specific heat of steam = 1850 J/kg/.

L = laten heat of vaporisation =

S = specific heat of water =

Li = Latent heat of freezing =

a) Now, lets assume the final temperature is T2 = 100 .

Therefore, heat lost is .

As E1 < E, Now, it change phase.

Let assume complete steam changes phase.

Then heat lost is .

As the sum of E1 and E2 is still less than E.

The water which is now at 100 degree lowers it's temperature.

Lets assume, it reaches to 0.

Then heat lost is .

As the sum of E1, E2 and E3 is still less than E, it will further looses heat and changes it's phase.

Let's assume total water is converted into ice.

Then heat lost is .

As the sum of E1, E2, E3 and E4 is greater than E, it means total water did not converted into ice.

Therefore, we have some water and some ice at 0.

Therefore, final temperature is 0. [answer]

b) The final phase is water-ice at 0. [answer]

c) Let M mass of water converted into ice.

Therefore, E1+E2+E3+MLi =Total heat removed

Therefore, mass of water converted into ice is 18.74 kg [answer]