An office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20.°C to 100. °C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume the wire is at 100. °C throughout the 4.00-min time interval. (a) Calculate the average power required to warm the water to 100. °C in 4.00 min. (b) Calculate the required resistance in the heating element at 100. °C. (c) Calculate the resistance of the heating element at 20. °C. (d) Derive a relationship between the diameter of the wire, the resistivity at 20. °C, ρ 0 , the resistance at 20. °C, R 0 , and the length L . (e) If L = 3.00 m, what is the diameter of the wire?

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Answer 1

Textbook Question

Chapter 17, Problem 48P

An office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20.°C to 100. °C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume the wire is at 100. °C throughout the 4.00-min time interval. (a) Calculate the average power required to warm the water to 100. °C in 4.00 min. (b) Calculate the required resistance in the heating element at 100. °C. (c) Calculate the resistance of the heating element at 20. °C. (d) Derive a relationship between the diameter of the wire, the resistivity at 20. °C, ρ₀, the resistance at 20. °C, R₀, and the length L. (e) If L = 3.00 m, what is the diameter of the wire?

(a)

Expert Solution

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

(b)

Expert Solution

To determine

The resistance of nichrome at

$100.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

(c)

Expert Solution

To determine

The resistance of nichrome at

$20.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

(d)

Expert Solution

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

(e)

Expert Solution

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2ρ0LπR0$

Substitute $150×10−8 Ω⋅m$ for $ρ0$ , $3.00 m$ for $L$ , $3.14$ for $π$ and $40 Ω$ for $R0$ in the above equation to find $d$

$d=2(150×10−8 Ω⋅m)(3.00 m)(3.14)(40 Ω)=3.78×10−4 m$

Conclusion: Therefore, the diameter of wire is $3.78×10−4 m$

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Answer 2

Textbook Question

Chapter 17, Problem 48P

An office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20.°C to 100. °C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume the wire is at 100. °C throughout the 4.00-min time interval. (a) Calculate the average power required to warm the water to 100. °C in 4.00 min. (b) Calculate the required resistance in the heating element at 100. °C. (c) Calculate the resistance of the heating element at 20. °C. (d) Derive a relationship between the diameter of the wire, the resistivity at 20. °C, ρ₀, the resistance at 20. °C, R₀, and the length L. (e) If L = 3.00 m, what is the diameter of the wire?

(a)

Expert Solution

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

(b)

Expert Solution

To determine

The resistance of nichrome at

$100.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

(c)

Expert Solution

To determine

The resistance of nichrome at

$20.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

(d)

Expert Solution

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

(e)

Expert Solution

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2ρ0LπR0$

Substitute $150×10−8 Ω⋅m$ for $ρ0$ , $3.00 m$ for $L$ , $3.14$ for $π$ and $40 Ω$ for $R0$ in the above equation to find $d$

$d=2(150×10−8 Ω⋅m)(3.00 m)(3.14)(40 Ω)=3.78×10−4 m$

Conclusion: Therefore, the diameter of wire is $3.78×10−4 m$

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Answer 3

Textbook Question

Chapter 17, Problem 48P

An office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20.°C to 100. °C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume the wire is at 100. °C throughout the 4.00-min time interval. (a) Calculate the average power required to warm the water to 100. °C in 4.00 min. (b) Calculate the required resistance in the heating element at 100. °C. (c) Calculate the resistance of the heating element at 20. °C. (d) Derive a relationship between the diameter of the wire, the resistivity at 20. °C, ρ₀, the resistance at 20. °C, R₀, and the length L. (e) If L = 3.00 m, what is the diameter of the wire?

(a)

Expert Solution

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

(b)

Expert Solution

To determine

The resistance of nichrome at

$100.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

(c)

Expert Solution

To determine

The resistance of nichrome at

$20.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

(d)

Expert Solution

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

(e)

Expert Solution

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2ρ0LπR0$

Substitute $150×10−8 Ω⋅m$ for $ρ0$ , $3.00 m$ for $L$ , $3.14$ for $π$ and $40 Ω$ for $R0$ in the above equation to find $d$

$d=2(150×10−8 Ω⋅m)(3.00 m)(3.14)(40 Ω)=3.78×10−4 m$

Conclusion: Therefore, the diameter of wire is $3.78×10−4 m$

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Answer 4

Textbook Question

Chapter 17, Problem 48P

An office worker uses an immersion heater to warm 250 g of water in a light, covered, insulated cup from 20.°C to 100. °C in 4.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume the wire is at 100. °C throughout the 4.00-min time interval. (a) Calculate the average power required to warm the water to 100. °C in 4.00 min. (b) Calculate the required resistance in the heating element at 100. °C. (c) Calculate the resistance of the heating element at 20. °C. (d) Derive a relationship between the diameter of the wire, the resistivity at 20. °C, ρ₀, the resistance at 20. °C, R₀, and the length L. (e) If L = 3.00 m, what is the diameter of the wire?

(a)

Expert Solution

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

(b)

Expert Solution

To determine

The resistance of nichrome at

$100.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

(c)

Expert Solution

To determine

The resistance of nichrome at

$20.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

(d)

Expert Solution

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

(e)

Expert Solution

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2ρ0LπR0$

Substitute $150×10−8 Ω⋅m$ for $ρ0$ , $3.00 m$ for $L$ , $3.14$ for $π$ and $40 Ω$ for $R0$ in the above equation to find $d$

$d=2(150×10−8 Ω⋅m)(3.00 m)(3.14)(40 Ω)=3.78×10−4 m$

Conclusion: Therefore, the diameter of wire is $3.78×10−4 m$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

(b)

Expert Solution

To determine

The resistance of nichrome at

$100.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

(c)

Expert Solution

To determine

The resistance of nichrome at

$20.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

(d)

Expert Solution

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

(e)

Expert Solution

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2ρ0LπR0$

Substitute $150×10−8 Ω⋅m$ for $ρ0$ , $3.00 m$ for $L$ , $3.14$ for $π$ and $40 Ω$ for $R0$ in the above equation to find $d$

$d=2(150×10−8 Ω⋅m)(3.00 m)(3.14)(40 Ω)=3.78×10−4 m$

Conclusion: Therefore, the diameter of wire is $3.78×10−4 m$

Answer 8

(a)

Expert Solution

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The average power required to warm the water.

Answer 13

Answer to Problem 48P

The average power required to warm the water is

$348.83 W$ .

Answer 14

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply, $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the energy required to warm the water is,

$E=mc(T2−T1)$

- $E$ is the heat energy required to warm the water from temperature $T1$ to $T2$ ,
- $m$ is the mass of water,
- $c$ is the specific heat capacity of water,

Substitute $250.0 g$ for $m$ , $4186 J(kg°C)−1$ for $c$ , $100.0 °C$ for $T2$ and $20.0 °C$ for $T1$ in the above equation to find $E$

$E=(250.0 g)(1 kg1000 g)(4186 J(kg°C)-1)(100.0 °C−20.0 °C)=83720 J(1 kJ1000 J)=83.72 kJ$

The heat energy required to warm the water from $20.0 °C$ to $100.0 °C$ in 4 minutes is $83.72 kJ$

Formula to calculate the average power required to warm the water is,

$P=EΔt$

- $P$ is the power required,
- $Δt$ is the time taken for warming the water,

Substitute $83.72 kJ$ for $E$ and 4 minutes for $Δt$ in the above equation to find $P$ ,

$P=83.72 kJ(103 J1 kJ)(4 minutes)(60 s1 minute)=348.83 W$

Conclusion: Therefore, the average power required to warm the water is $348.83 W$

Answer 15

(b)

Expert Solution

To determine

The resistance of nichrome at

$100.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The resistance of nichrome at

$100.0 °C$

Answer 20

Answer to Problem 48P

The resistance of nichrome at

$100.0 °C$ is

$41.28 Ω$

Answer 21

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply

Explanation:

Formula to calculate the resistance is,

$R=(ΔV)2P$

- $R$ is the resistance of nichrome at $100.0 °C$
- $ΔV$ is the voltage of power supply,

Substitute $120 V$ for $ΔV$ and $348.83 W$ for $P$ in the above equation to find $R$ ,

$R=(120 V)2348.83 W=41.28 Ω$

Conclusion: Therefore, the resistance of nichrome at $100.0 °C$ is $41.28 Ω$

Answer 22

(c)

Expert Solution

To determine

The resistance of nichrome at

$20.0 °C$

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The resistance of nichrome at

$20.0 °C$

Answer 27

Answer to Problem 48P

The resistance of nichrome at

$20.0 °C$ is

$40 Ω$

Answer 28

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance is,

$R0=R1+α(T−T0)$

- $R0$ is the resistance at $T0$ ,
- $α$ is the temperature coefficient of resistivity of nichrome,
- $T$ is the final temperature,

Substitute $41.28 Ω$ for $R$ , $0.4×10−3 (°C)-1$ for $α$ , $100.0 °C$ for $T$ and $20.0 °C$ for $T0$ in the above equation to find $R$

$R0=41.28 Ω(1+(0.4×10−3 (°C)-1)(100.0 °C−20.0 °C))=40 Ω$

Conclusion: Therefore, the resistance of nichrome at $20.0 °C$ is $40 Ω$

Answer 29

(d)

Expert Solution

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The relationship between resistivity at

$20.0 °C$ , diameter of wire, resistance at

$20.0 °C$ and length of wire.

Answer 34

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is

$d=2ρ0LπR0$

Answer 35

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply.

Explanation:

Formula to calculate the resistance at $20.0 °C$ is,

$R0=ρ0LA$

- $ρ0$ is the resistivity of nichrome wire at $20.0 °C$ ,
- $L$ is the length of nichrome wire,
- $A$ is the area of circular cross section of nichrome wire,

Equation to calculate the area of cross section is ,

$A=πd24$

- $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R0$

$R0=4ρ0Lπd2$

Rewrite the above equation in terms of $d$

$d=2ρ0LπR0$

Is the equation which relates diameter resistance, resistivity and length of wire,

Conclusion: The equation which relates diameter resistance, resistivity and length of wire is $d=2ρ0LπR0$

Answer 36

(e)

Expert Solution

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2ρ0LπR0$

Substitute $150×10−8 Ω⋅m$ for $ρ0$ , $3.00 m$ for $L$ , $3.14$ for $π$ and $40 Ω$ for $R0$ in the above equation to find $d$

$d=2(150×10−8 Ω⋅m)(3.00 m)(3.14)(40 Ω)=3.78×10−4 m$

Conclusion: Therefore, the diameter of wire is $3.78×10−4 m$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The diameter of nichrome wire.

Answer 41

Answer to Problem 48P

The diameter of nichrome wire is

$3.78×10−4 m$

Answer 42

Explanation of Solution

Given Info: $250.0 g$ of water is heated from $20.0 °C$ to $100.0 °C$ in 4 minutes. The heater with nichrome resistor is connected to $120 V$ supply. The length of nichrome wire is $3.00 m$