Question

A 200g 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. a) find the final temperature of the system, assuming no heat losses. (Assume that the specific heat of ice is 2.0 kj/kg K) b) A second 200 g piece of ice at -20 C is added. How much ice remains in the system after it reaches equilibrium? c) would your answer in 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

Here,

a) let the final temperature of the system is T

heat needed to melt all the ice

Q = 100 * 334 + 100 * 2 * 20 = 37400 J

heat lost by Al and water = 200 * 0.9 * 20 + 500 * 4.186 * 20

heat lost by Al and water = 45460 J

hence , all the ice will melt

37400 + 100 * 4.186 * (T - 0) = 200 * 0.90 * (20 - T) + 500* 4.186 * (20 - T)

solving for T

T = 3 degree C

the final temperature is 3⁰ C

b) let the mass of ice remains is x

heat lost = heat gain by ice

45460 = 300 * 2 * (20) + (300 - x) * 334

solving for x

x = 199.2 gm

the mass of ice remaining is 199.2 gm

c) no still the answer will remain the same