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

Question

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Answer 1

Question

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

Answer to Problem 41P

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

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

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Answer 2

Question

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

Answer to Problem 41P

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

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

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Answer 3

Question

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

Answer to Problem 41P

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

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

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Answer 4

Question

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

Answer to Problem 41P

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

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

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Answer 5

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

Answer to Problem 41P

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

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

Answer 6

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

Answer to Problem 41P

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

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

Answer 7

Chapter 14, Problem 41P

To determine

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

Answer 8

Expert Solution & Answer

Answer to Problem 41P

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

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The temperature of the ice is $−10.0 °C$ . The volume of tea is $2.00×10−4 m3$ . 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.

$Qt=ρwVtcw(Tf−Tti)$ (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).

$miciΔT1+miL+micwΔT2+ρwVtcw(Tf−Tti)=0$

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

$mi(ciΔT1+L+cwΔT2)=−ρwVtcw(Tf−Tti)mi=−ρwVtcw(Tf−Tti)ciΔT1+L+cwΔT2$ (IV)

Conclusion:

Substitute $2.00×10−4 m3$ for $Vt$ , $95.0 °C$ for $Tti$ , $10.0 °C$ for $Tf$ , $0.0 °C−(−10.0 °C)$ for $ΔT1$ , $10.0 °C−0.0 °C$ for $ΔT2$ , $1000 kg/m3$ for $ρw$ , $4.186 kJ/kg⋅K$ for $cw$ , $2.1 kJ/kg⋅K$ for $ci$ , $333.7 kJ/kg$ for $L$ .

$mi=−(1000 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K)(10.0 °C−95.0 °C)(4.186 kJ/kg⋅K)(0.0 °C−(−10.0 °C))+333.7 kJ/kg+(10.0 °C−0.0 °C)(2.1 kJ/kg⋅K)=179 g$

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

Answer 12

Ch. 14.1 - 14.1 On the Rebound If a rubber ball of mass 1.0...Ch. 14.1 - Prob. 14.2PP Ch. 14.2 - CHECKPOINT 14.2 Take a rubber band and stretch it...Ch. 14.2 - Prob. 14.3PP Ch. 14.3 - Prob. 14.3CP Ch. 14.3 - Prob. 14.4PP Ch. 14.3 - Prob. 14.5PP Ch. 14.4 - Prob. 14.6PP Ch. 14.5 - Prob. 14.5CP Ch. 14.5 - Prob. 14.7PP

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

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

Answer to Problem 41P

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Chapter 14 Solutions

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

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How much ice should be added to the glass of tea to cool it to a temperature of 10.0 °C<m:math><m:mrow><m:mn>10.0</m:mn><m:mtext> </m:mtext><m:mtext>°C</m:mtext></m:mrow></m:math>.

Answer to Problem 41P

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