A block of ice of mass 0.8 kg and initial temperature T = 0 o C is placed in a s

Question

A block of ice of mass 0.8 kg and initial temperature T = 0 oC is placed in a sealed insulated container full of Helium gas, initially at temperature 190 oC and pressure of 1 atm.   The volume of the Helium is 870 L, and is constant. Helium is a monatomic ideal gas.

What is the mass of the liquid water when the system comes to equilibrium? (In other words, how much ice melts?)

Assume no heat is lost to the surroundings.

Give your answer in kg to three significant digits. Do not include units in your answer.

Explanation / Answer

given

T = 190 + 273 = 463 k

V = 870 L = 0.87 m^3

P = 1 atm = 1.013*10^5 pa

Let n is the no of moles of the gas.

use, P*V = n*R*T

==> n = P*V/(R*T)

= 1.013*10^5*0.87/(8.314*463)

= 22.89 mol

here final temperarure of helium will be 0C

let m is the mass of ice that melts.

now Apply, Heat lost by gained ice = heat lost by helium

m*Lf = (3/2)*n*R*dT

m = (3/2)*n*R*dT/Lf

= (3/2)*22.89*8.314*( 190 - 0)/(3.33*10^5)

= 0.1628 kg <<<<<<<------------Answer

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