Question

Suppose you have 600.0 gramsof room temperature water (20.0 degrees Celsius) in a thermos. You drop 90.0 grams of ice at 0.00 degrees Celsius into the thermos and shut the lid.

(a) What is the equilibrium temperature of the system?

(b) How much ice is left (in grams)?

Answer 1

Step 1

Given : Mass of water = 600.0 g

Initial temperature of water = 20.0 ^oC

Mass of ice = 90.0 g

And initial temperature of ice added = 0.0 ^oC

 

Step 2

Assuming the final temperature solution at equilibrium is T and all the ice has melted.

The total amount of energy lost by the water added is given by,

=> Q = mCΔT

where m = mass of water added = 600.0 g

C = specific heat of water = 4.184 J/g.^oC

And ΔT = change in temperature of water = T - 20 = T -20 ^oC

Hence substituting the values we get,

=> Q = 600.0 X 4.184 X (T - 20) = 2510.4 T - 50208 

Since all the heat lost by water is gained by the ice.

Hence the heat gained by ice = -(heat lost by water) = 50208 - 2510.4 T                   (-ve because heat is lost by water)

Step 3

  The total amount of energy gained by the ice is given by,

=> Q = mΔH + mCΔT

where m = mass of ice = 90.0 g

C = specific heat of water = 4.184 J/g.^oC

ΔT = change in temperature of melted water = T - 0.0 = T 

And ΔH = heat of vaporisation of ice = 334 J/g.

Hence substituting the values we get,

=> 50208 - 2510.4 T = 90.0 X 334 + 90.0 X 4.184 X T

=> T = 6.98 ^oC

Hence the equilibrium temperature the system is 6.98 ^oC.

And since as assumed initially, the final temperature is coming more than 0.0 ^oC. Hence all the ice has melted.

Hence mass of mass of ice left is 0.0 g.

Given : Mass of water = 600.0 g

Initial temperature of water = 20.0 ^oC

Mass of ice = 90.0 g

And initial temperature of ice added = 0.0 ^oC