Question

7.1 g of water ice at 268.4 K is mixed with 234 g of liquid water at 304.7 K in an insulated container under an external pressure of 1.00 bar. Answer the following questions assuming that the physical properties of water and ice are constant over the range of interest. Physical properties for water and ice are given below:

Cp,m,ice = 1.95 J g-1 K-1
ΔHfusion,ice = 333.4 J g-1
Cp,m,water = 4.18 J g-1 K-1
Melting point of ice: 273.15 K

1) What will be the final form of the system? Select one that applies

(a)
Water

(b)
Mixed water and ice

(c)
Ice

2)What is the final temperature, Tf (in K), of the system when it reaches equilibrium?
Tf = ____________ K.

Answer 1

1. a) water

2. 301.3 K

Explanation:

The calculation is done assuming that the final temperature would be such that it will melt the 7.1 g of water ice at 268.4 K and increase the temperature of the resulting water as the 234 g of liquid water at 304.7 K is mixed to the water ice. This assumption is valid as can be seen from the value of final temperature.

Final temperature when the system reaches equilibrium = T_f K

Mass of ice = 7.1 g

Temperature of ice (T_ice) = 268.4 K

Specific heat of ice, (C_p,m) = 1.95 J g^-1 K^-1

Heat absorbed by ice to reach 273.15 K = mass of ice x specific heat of ice x (273.15- T_ice)

= 7.1 g x 1.95 J g^-1 K^-1 x(273.15 - 268.4) K

= 65.76 J

Heat absorbed on melting of ice = mass of ice x ΔH_fusion,ice

= 7.1 g x 333.4 J g^-1

= 2367 J

Heat absorbed by the melted ice (i.e. water) = mass of melted ice (i.e. water) x specific heat of water x (T_f- 273.15 K)

= 7.1 g x 4.18 J g^-1 K^-1 x (T_f- 273.15) K

= 29.68 x (T_f- 273.15) J

Mass of water mixed = 234 g

Temperature of water (T_water) = 304.7 K

Specific heat of water, (C_p,m) = 4.18 J g^-1 K^-1

Heat released by water = mass of water x specific heat of water x (T_f - T_water)

= 234 g x 4.18 J g^-1 K^-1 x (T_f - 304.7) K

= 978.1 x (T_f - 304.7) J

Now,

total heat absorbed + total heat released = 0

or, 65.76 J +  2367 J + 29.68 x (T_f- 273.15) J + 978.1 x (T_f - 304.7) J = 0

or, 2433 J + 29.68T_f J - 8107 J + 978.1T_f J - 298027 J = 0

or, 1008T_f = 303701

or, T_f = 301.3 K

Thus, the final temperature (T_f) of the system when it reaches equilibrium = 301.3 K

At this temperature of 301.3 K, the final form of the system is (a) Water.