Suppose that in an insulated container, 0.100 kg of water at 20.0C is mixed with

Question

Suppose that in an insulated container, 0.100 kg of water at 20.0C is mixed with 1.500kg of ice at 15.0C. You are asked to find the final temperature Tf of the system, but you are not told what the final phase of the system in equilibrium is.

PART A) Find the amount of heat Qh it would take to bring all of the ice to its melting point, 0C.

Express your answer in joules to three significant figures.

PART B)

Find the amount of heat Qc released in bringing all of the water to its freezing point, 0C. Your answer does not need to include a sign.

Express your answer in joules to three significant figures.

Explanation / Answer

This is a problem calorimetry, we will use two expressions

Q = m Ce (Tf- Ti)

Q = m L for stated change

Part A)

Data

m = 1.500 Kg

Ti = -15.0 C

Ce =2090 J/ KgC

Q = 1.5 2090 ( 0-(-15) )

Qi = 4.70 104 J

This is the heat to bring the ice to 0C unchanged stated

Part B)

Data

m = 0.100 Kg

Ti = 20.0 C

Ce = 4186 J/Kg C

Q = 0.1 4186 (20 – 0)

Qw = 8.37 103 J

This is the heat to bring water to 0C unchanged stated

Part C)

We will find the final temperature of the system

Qi – Qw > 0

this result we can say that all the water is zero degrees and the ice has not melted yet, so the system in its final state is in the solid state.

Let's calculate the heat released by freezing water

Qw= 8.37 103J

The heat necessary for the state change L = 5.23 103 J/Kg

Q2 = 0.100 5.23 103J

Q2 = 0.523 103J

the total heat required to transform all water is ice to 0C

Qt =Qw + Q2

Qt =(8.37 + 0.523 ) 103 J

Qt = 8.893 103 J

This heat is transferred to the ice mass that is initially Ti = -15C up to temperature

Qt = mi Cei (Tf-Ti)i

Tf = Ti + Qt/miCei

Tf = -15 + 8.893 103 /1.5 2090

Tf = -15 + 2.837

Tf = -12.16C

With this calculation we have the two masses of ice that should lead to an equilibrium temperature

m1= 1.5 Kg

T1 = -12.16 C

m.2 = 0.1 Kg

T.2 = 0 C

Qc = Qa

m1 Ce ( Tf -T1) = m.2 Ce ( T.2 - Tf)

m1/m.2 (Tf-T1)= ( T.2 - Tf)

1.5/0.1 (Tf +12.16) = ( 0 -Tf)

15 Tf + 15 12.16 = -Tf

16 Tf = -182.4

Tf = -11.4 C

This is the final temperature of the ice mass

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