The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ / kg ⋅ K .

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Answer 1

Question

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.

Substitute 242 g for mice and 179 g for mi in the equation for Δ.

Δ=(242 g−179 g179 g)100%=35%

Therefore, the percentage change is 35%.

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Answer 2

Question

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.

Substitute 242 g for mice and 179 g for mi in the equation for Δ.

Δ=(242 g−179 g179 g)100%=35%

Therefore, the percentage change is 35%.

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Answer 3

Question

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.

Substitute 242 g for mice and 179 g for mi in the equation for Δ.

Δ=(242 g−179 g179 g)100%=35%

Therefore, the percentage change is 35%.

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Answer 4

Question

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.

Substitute 242 g for mice and 179 g for mi in the equation for Δ.

Δ=(242 g−179 g179 g)100%=35%

Therefore, the percentage change is 35%.

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Answer 5

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.

Substitute 242 g for mice and 179 g for mi in the equation for Δ.

Δ=(242 g−179 g179 g)100%=35%

Therefore, the percentage change is 35%.

Answer 6

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.

Substitute 242 g for mice and 179 g for mi in the equation for Δ.

Δ=(242 g−179 g179 g)100%=35%

Therefore, the percentage change is 35%.

Answer 7

Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass 350 g and specific heat 0.837 kJ/kg⋅K.

Answer 8

Expert Solution & Answer

Answer to Problem 42P

The percentage change is 35%.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

Qt+Qice+Qg=0 (I)

Here, Qt is the heat energy possessed by tea, Qice is the heat energy possessed by ice, and Qg is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for Qt.

Qt=ρwVtcw(Tf−Ti) (II)

Here, ρw is the density of water, Vt of tea, cw is the specific heat of water,Ti is the initial temperature of tea and glass, and Tf is the final temperature of tea and glass.

Write the equation for Qice.

Qice=miceLf+micecice(T1f−T1i)+micecw(T2f−T2i) (III)

Here, mice is the mass of ice, Lf is the latent heat of fusion of ice, cice is the specific heat of ice, T1i is the initial temperature of ice, T1f is the final temperature of ice, temperature change of ice,T2f is the final temperature of ice-water, and T2i is the final temperature of ice-water.

Write the equation for Qg.

Qg=mgcg(Tf−Ti) (IV)

Here, mg is the mass of glass and cg is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

ρwVtcw(Tf−Ti)+[miceLf+micecice(T1f−T1i)+micecw(T2f−T2i)]+mgcg(Tf−Ti)=0

Rewrite the above relation in terms of mice.

mice=−(ρwVtcw+mgcg)(Tf−Ti)Lf+cice(T1f−T1i)+cw(T2f−T2i)

Write the equation to find the percentage change.

Δ=(mice−mimi)100%

Here,Δ is the percentage change and mi is the mass of ice calculated in problem 41.

Conclusion:

Substitute 1.00×103 kg/m3 for ρw, 2.00×10−4 m3 for Vt, 4.186 kJ/kg⋅K for cw, 0.35 kg for mg, 0.837 kJ/kg⋅K for cg, 10.0°C for Tf, 95.0°C for Ti, 333.7 kJ/kg for Lf, 2.1 kJ/kg⋅K for cice, −10.0°C for T1i , 0°C for T1f, 0°C for T2i, and 10.0°C for T2f in the equation for mice.

mice=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))(10°C−(95.0°C))(333.7 kJ/kg+((2.1 kJ/kg⋅K)(0°C−(−10.0°C)))+(4.186 kJ/kg⋅K)(10.0°C−(0°C)))=−(((1.00×103 kg/m3)(2.00×10−4 m3)(4.186 kJ/kg⋅K))+((0.35 kg)(0.837 kJ/kg⋅K)))((10+273 )K−(95.0+273) K)(333.7 kJ/kg+((2.1 kJ/kg⋅K)[(0+273 ) K−(−10.0+273 ) K])+(4.186 kJ/kg⋅K)[(10.0+273 ) K−(0+273 ) K])=(837.2 J/K+292.95 J/K)(85 K)(354.7 kJ/kg+41.86 kJ/kg)=0.242 kg(103 g1 kg)=242 g.