Question

Suppose you wanted to cool 100. g of water from 20 C to 0 C using dry ice, CO2 (s). The enthalpy of sublimation is 25.2 kJ/mol. What mass of dry ice should you need? (a) 0.33 g (b) 15 g (c) 3.5 g (d) 150 g ----- The correct answer is b, I'm just unsure on how to get there.

Answer 1

Given :

Mass of water = 100.0 g

Initial T of water = 20 degree , Final T = 0 degree C

Enthalpy of sublimation of dry ice = 25.2 kJ /mol

Solution :

Here we have to calculated mass of dry ice needed to require to lower the temperature of water from 20 degree C to 0 degree C

Let’s calculate the heat released in cooling the water from 20 deg C to 0 deg C

The formula is

q = m C delta T

here heat released in J , m is mass in g , C is sp heat of water in J / g deg C , Delta T is change in T.

Lets use given values and get q .

q = 100.0 g x 4.184 J / g deg C x ( 0 deg C – 20 .0 deg C )

= - 8368 J

We now have to get the mass of dry ice.

Lets first find out the moles of dry ice

n CO2 = 8368 J / heat of sublimation

= 8368 J / 25.2 E 3 J per mol

= 0.332 mol

Moles of CO2 = 0.332 mol

Mass of CO2 in g = 0.332 mol x molar mass of CO2

= 0.332 mol CO2 x 44.01 g /mol

= 14.61 g

Mass of CO2 = 14.61 g