Question

1. A 200 g aluminum calorimeter can contain 500 g of water at 20 C. A 100 g piece of ice cooled to -20 C is placed in the calorimeter.

1. Find the final temperature of the system, assuming no heat losses. (Assume that the specific heat of ice is 2.0 kJ/kg K)
2. A second 200 g piece of ice at -20 C is added. How much ice remains in the system after it reaches equilibrium?
3. Would your answer to part b be different if both pieces of ice were added at the same time?

(specific heat of aluminum is 0.9 kJ/kg K and water is 4.18 kJ/kg K)

Answer 1

mass of alumnium , m1 = 200 g

mass of water , m2 = 500 g

mass of ice , m3 = 100 g

a)

let the final temperature be Tf

using conservation of heat energy

heat gained by ice = heat lost by water and alumunium

m3 * ( Ci * ( 0 - (-20)) + Lf + Cw * ( Tf - 0)) = m1 * Ca * ( 20 - Tf) + m2 * Cw * ( 20 - Tf)

100 * ( 2 * ( 0 - (-20)) + 334 + 4.186 * ( Tf - 0)) = 200 * 0.9 * ( 20 - Tf) + 500 * 4.186 * ( 20 - Tf)

solving for Tf

Tf = 3 degree

b)

the heat required by water to attain final temperature 0 degree C , Q1 = (m2 + m3) * Cw * ( Tf - 0) + m1 * Ca * ( Tf - 0)

Q1 = (500 + 100) * 4.186 * ( 3) + 200 * 0.9 * ( 3 - 0)

Q1 = 8000 J

the heat required for ice block to be at 0 degree C , Q2= 200 * 2 * ( 0 - (-20)) = 8000 J

as Q1 = Q2

no ice will be melted

the mass of ice remained in the system is 200 g

c)

as the answer of parts does not depends on time

Yes , the answer will be same