A 190 g piece of ice at 0°C is placed in 400 g of water at...

A 190 g piece of ice at 0°C is placed in 400 g of water at 24°C. The system is in a container of negligible heat capacity and is insulated from its surroundings.

(a) What is the final equilibrium temperature of the system?
°C

(b) How much of the ice melts?
g

Solutions

Expert Solution

Unless asked to solve using SI units, it is much easier tosolve in cgs units and calorie for heat.

If m gram of ice melts and Tf is the final temp the heatgained by ice & the melted water = Heat lost by the warmwater.

Then Q1 = Q2

gives, m.L + mct = Mc(24-t)

Two unknowns and one eqn only. Difficult to solve.

Let us see if all of the ice is melted.

Heat required to melt the 190 g of ice,

Q = mL = 190x76 = 14440 cal.

Heat available in all of the warm water cooling down to0^oC,

Q' = 400x1x24 = 9600 cal

Since Q> Q' allof the ice cannot melt.

So, the warm water will cool down to 0oC melting some of theice, say m' gram.

Heat gained by the ice melted, Q" = m.L = 76 m cal.

Heat lost by the warm water, Q' = 400x24 = 9600 cal.

These are equal.Q'' = Q'

76 m = 9600

m = 9600/76 = 126.32 g

Final temp is ofcourse, 0^oC, as some ice isstill not melted.

genius_generous answered 1 year ago

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