Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 14, Problem 48P

(a)

To determine

The amount of ice need to be put in 0.250kg of water at 25.0°C to reduce the temperature of the water to 0°C.

(a)

Expert Solution
Check Mark

Answer to Problem 48P

The mass of ice need to be added is 74g.

Explanation of Solution

The temperature of the ice is 10.0°C. The temperature of water is 25.0°C. Mass of water is 0.250kg. The final temperature is 0°C.

The change in internal energy of the ice-water system is zero.

Write the equation for the first law of thermodynamic for the system.

Qw+Qi=ΔU=0 (I)

Here, ΔU is the change in internal energy, Qw is the heat of the water, Qi is the ice.

The temperature of the water changes from 25.0°C to 0°C.

Write the equation for the heat change of the water.

Qw=mwcw(TfTwi) (II)

Here, mw is the mass of the water, cw is the specific heat of water, Tf is the final temperature, Twi is the initial temperature of water.

The temperature of the ice first changes from 10.0°C to 0.0°C, then ice turns to water at 0.0°C.

Write the equation for the heat change of the ice.

Qi=miciΔT1+miL (III)

Here, mi is the mass of ice, ci is the specific heat of ice, ΔT1 is the change in temperature from 10.0°C to 0.0°C, L is the latent heat of fusion of water.

Substitute equation (II) and (III) in equation (I).

miciΔT1+miL+mwcw(TfTwi)=0

Re-write the above equation to get an expression for mi.

mi(ciΔT1+L)=mwcw(TfTwi)mi=mwcw(TfTwi)ciΔT1+L (IV)

Conclusion:

Substitute 0.250kg for mw, 25.0°C for Twi, 0.0°C for Tf, 0.0°C(10.0°C) for ΔT1, 4.186kJ/kgK for cw, 2.1kJ/kgK for ci, 333.7kJ/kg for L.

mi=(0.250kg)(4.186kJ/kgK)(0.0°C25.0°C)(2.1kJ/kgK)(0.0°C(10.0°C))+333.7kJ/kg=74g

The mass of ice need to be added is 74g.

(b)

To determine

The final temperature if half of the ice as section (a) is added to the water.

(b)

Expert Solution
Check Mark

Answer to Problem 48P

The final temperature is 11°C.

Explanation of Solution

The temperature of the ice is 10.0°C. The temperature of water is 25.0°C. Mass of water is 0.250kg. The final temperature is 0°C.

The change in internal energy of the ice-water system is zero.

Write the equation for the first law of thermodynamic for the system.

Qw+Qi=ΔU=0 (I)

Here, ΔU is the change in internal energy, Qw is the heat of the water, Qi is the ice.

The temperature of the water changes from 25.0°C to 0°C.

Write the equation for the heat change of the tea.

Qw=mwcw(TfTwi) (II)

Here, mw is the mass of the water, cw is the specific heat of water, Tf is the final temperature, Twi is the initial temperature of water.

The temperature of the ice first changes from 10.0°C to 0.0°C, then ice turns to water at 0.0°C, then the temperature of the water changes from 0.0°C to the final temperature.

Write the equation for the heat change of the ice.

Qi=mici(Tii0°C)+miL+micw(T0°C) (III)

Here, mi is the mass of ice, ci is the specific heat of ice, Tii is the initial temperature of the ice, L is the latent heat of fusion of water.

Substitute equation (II) and (III) in equation (I).

mici(Tii0°C)+miL+micw(T0°C)+mwcw(TfTwi)=0

Re-write the above equation to get an expression for Tf.

Tfcw(mw+mi)=mici(Tii0°C)miL+micw(0°C)+mwcwTwiTf=mici(Tii0°C)miL+micw(0°C)+mwcwTwicw(mw+mi)

Conclusion:

Substitute 0.250kg for mw, 25.0°C for Twi, 74g/2 for mi, 4.186kJ/kgK for cw, 2.1kJ/kgK for ci, 333.7kJ/kg for L, 10.0°C for Tii.

Tf=[(74g/2)(2.1kJ/kgK)(10.0°C0°C)(74g/2)(333.7kJ/kg)+(74g/2)(4.186kJ/kgK)(0°C)+(0.250kg)(4.186kJ/kgK)(25.0°C)](4.186kJ/kgK)(0.250kg+74g/2)=[(74g/2)(2.1kJ/kgK)(10.0+273K0+273K)(74g/2)(333.7kJ/kg)+(74g/2)(4.186kJ/kgK)(0+273K)+(0.250kg)(4.186kJ/kgK)(25.0+273K)](4.186kJ/kgK)(0.250kg+74g/2)=284K=284273°C=11°C

The final temperature is 11°C.

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Chapter 14 Solutions

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