Answered: In an insulated vessel, 250 g of ice at…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter17: Energy In Thermal Processes: The First Law Of Thermodynamics

Section: Chapter Questions

Problem 11P

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Question

In an insulated vessel, 250 g of ice at 0°C is added to 600 g of water at 18.0°C.
(a) What is the final temperature of the system?
(b) How much ice remains when the system reaches equilibrium?

Expert Solution

Step 1

Given that

Mass of the ice is m₁ = 250 g

Initial temperature of ice = 0 $° C$

Mass of the water m₂ = 600 g

Initial temperature of water is = $18 ° C$

Constant material depended parameters are

Specific heat of water s = 1 cal/g/ $° C$

Latent heat melting of ice is L = 80 cal/g

Step 2

Part a:

The heat will supply to the ice by the water.

The heat released by the water in the conversion from $18 ° C t o 0 ° C$ .

$Q_{1} = m_{2} s ∆ θ Q_{1} = (600) (1) (18 - 0) Q_{1} = 10800 c a l$

Now the amount of Ice melt by absorbed heat 10800 cal from water is,

$m = \frac{Q_{1}}{L} m = \frac{10800}{80} m = 135 g$

Now after this conversion

Ice of temperature 0 $° C$ remains in the system = (250-135) = 115 g .................(1)

Water of temperature $0 ° C$ remains in the system = (600+135) = 735 g

Now in the system, Ice of $0 ° C$ and water of $0 ° C$ temperature are present so the heat transfer from water to ice make same amount of ice and water in the system.

$H_{2} O_{(l i q u i d)} ⇋ H_{2} O_{(s o l i d)}$

Above equation is at constant equilibrium temperature of $0 ° C$

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