Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 14, Problem 41P
To determine

How much ice should be added to the glass of tea to cool it to a temperature of 10.0°C.

Expert Solution & Answer
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Answer to Problem 41P

The mass of ice need to be added is 179g.

Explanation of Solution

The temperature of the ice is 10.0°C. The volume of tea is 2.00×104m3. The initial temperature of hot tea is 95.0°C. The final temperature of tea is 10.0°C.

Since the container is insulated, the change in internal energy of the tea-ice system is zero.

Write the equation for the first law of thermodynamic for the system of ball and water.

Qt+Qi=ΔU=0 (I)

Here, ΔU is the change in internal energy, Qt is the heat of the tea, Qi is the ice.

The temperature of the tea changes from 95.0°C to 10.0°C.

Write the equation for the heat change of the tea.

Qt=ρwVtcw(TfTti) (II)

Here, ρw is the density of the water, Vt is the volume of tea, cw is the specific heat of water, Tf is the final temperature, Tti is the initial temperature of tea.

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 changed from 0.0°C to 10.0°C.

Write the equation for the heat change of the ice.

Qi=miciΔT1+miL+micwΔT2 (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, ΔT2 is the temperature change 0.0°C to 10.0°C.

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

miciΔT1+miL+micwΔT2+ρwVtcw(TfTti)=0

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

mi(ciΔT1+L+cwΔT2)=ρwVtcw(TfTti)mi=ρwVtcw(TfTti)ciΔT1+L+cwΔT2 (IV)

Conclusion:

Substitute 2.00×104m3 for Vt, 95.0°C for Tti, 10.0°C for Tf, 0.0°C(10.0°C) for ΔT1, 10.0°C0.0°C for ΔT2, 1000kg/m3 for ρw, 4.186kJ/kgK for cw, 2.1kJ/kgK for ci, 333.7kJ/kg for L.

mi=(1000kg/m3)(2.00×104m3)(4.186kJ/kgK)(10.0°C95.0°C)(4.186kJ/kgK)(0.0°C(10.0°C))+333.7kJ/kg+(10.0°C0.0°C)(2.1kJ/kgK)=179g

The mass of ice need to be added is 179g.

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