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?

To determine

The average power required to warm the water.

Answer to Problem 48P

The average power required to warm the water is $348.83\text{W}$.

Explanation of Solution

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

Explanation:

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

$E=mc\left(T_2-T_1\right)$

• • $E$ is the heat energy required to warm the water from temperature $T_1$ to $T_2$,
  • $m$ is the mass of water,
  • $c$ is the specific heat capacity of water,

Substitute $250.0\text{g}$ for $m$, $4186J{\left(kg°C\right)}^{-1}$ for $c$, $100.0\text{°C}$ for $T_2$ and $20.0\text{°C}$ for $T_1$ in the above equation to find $E$

$\begin{array}{c}E=\left(250.0\text{g}\right)\left(\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\right)\left(4186\text{J}{\left(\text{kg°C}\right)}^{\text{-1}}\right)\left(100.0\text{°C}-20.0\text{°C}\right)\\ =\text{83720}\text{J}\left(\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\right)\\ \text{=83}\text{.72}\text{kJ}\end{array}$

The heat energy required to warm the water from $20.0\text{°C}$ to $100.0\text{°C}$ in 4 minutes is $\text{83}\text{.72}\text{kJ}$

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

$P=\frac{E}{\Delta t}$

• • $P$ is the power required,
  • $\Delta t$ is the time taken for warming the water,

Substitute $\text{83}\text{.72}\text{kJ}$ for $E$ and 4 minutes for $\Delta t$ in the above equation to find $P$,

$\begin{array}{c}P=\frac{\text{83}\text{.72}\text{kJ}\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\right)}{\left(4\text{ minutes}\right)\left(\frac{\text{60}\text{s}}{1\text{minute}}\right)}\\ =348.83\text{W}\end{array}$

Conclusion: Therefore, the average power required to warm the water is $348.83\text{W}$

To determine

The resistance of nichrome at $100.0\text{°C}$

Answer to Problem 48P

The resistance of nichrome at $100.0\text{°C}$ is $41.28\Omega$

Explanation of Solution

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

Explanation:

Formula to calculate the resistance is,

$R=\frac{{\left(\Delta V\right)}^2}{P}$

• • $R$ is the resistance of nichrome at $100.0\text{°C}$
  • $\Delta V$ is the voltage of power supply,

Substitute $120\text{V}$ for $\Delta V$ and $348.83\text{W}$ for $P$ in the above equation to find $R$,

$\begin{array}{c}R=\frac{{\left(120\text{V}\right)}^2}{348.83\text{W}}\\ =41.28\Omega \end{array}$

Conclusion: Therefore, the resistance of nichrome at $100.0\text{°C}$ is $41.28\Omega$

To determine

The resistance of nichrome at $20.0\text{°C}$

Answer to Problem 48P

The resistance of nichrome at $20.0\text{°C}$ is $40\Omega$

Explanation of Solution

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

Explanation:

Formula to calculate the resistance is,

$R_0=\frac{R}{1+\alpha \left(T-T_0\right)}$

• • $R_0$ is the resistance at $T_0$,
  • $\alpha$ is the temperature coefficient of resistivity of nichrome,
  • $T$ is the final temperature,

Substitute $41.28\Omega$ for $R$, $0.4\times {10}^{-3}{\left(\text{°C}\right)}^{\text{-1}}$ for $\alpha$, $100.0\text{°C}$ for $T$ and $20.0\text{°C}$ for $T_0$ in the above equation to find $R$

$\begin{array}{c}{R}_0=\frac{41.28\Omega }{\left(1+\left(0.4\times {10}^{-3}{\left(\text{°C}\right)}^{\text{-1}}\right)\left(100.0\text{°C}-20.0\text{°C}\right)\right)}\\ =40\Omega \end{array}$

Conclusion: Therefore, the resistance of nichrome at $20.0\text{°C}$ is $40\Omega$

To determine

The relationship between resistivity at $20.0\text{°C}$, diameter of wire, resistance at $20.0\text{°C}$ and length of wire.

Answer to Problem 48P

The equation which relates diameter resistance, resistivity and length of wire is $d=2\sqrt{\frac{{\rho }_{0}L}{\pi R_0}}$

Explanation of Solution

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

Explanation:

Formula to calculate the resistance at $20.0\text{°C}$ is,

$R_0=\frac{{\rho }_{0}L}{A}$

• • ${\rho }_0$ is the resistivity of nichrome wire at $20.0\text{°C}$,
  • $L$ 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 ,

$A=\frac{\pi d^2}{4}$

• • $d$ is the diameter of wire,

Substitute the above equation in the previous equation to rewrite $R_0$

$R_0=\frac{4{\rho }_{0}L}{\pi d^2}$

Rewrite the above equation in terms of $d$

$d=2\sqrt{\frac{{\rho }_{0}L}{\pi R_0}}$

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 $d=2\sqrt{\frac{{\rho }_{0}L}{\pi R_0}}$

To determine

The diameter of nichrome wire.

Answer to Problem 48P

The diameter of nichrome wire is $3.78\times {10}^{-4}\text{m}$

Explanation of Solution

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

Explanation:

Formula to calculate the length of nichrome wire is,

$d=2\sqrt{\frac{{\rho }_{0}L}{\pi R_0}}$

Substitute $150\times {10}^{-8}\text{Ω}\cdot \text{m}$ for ${\rho }_0$, $3.00\text{m}$ for $L$, $3.14$ for $\pi$ and $40\Omega$ for $R_0$ in the above equation to find $d$

$\begin{array}{c}d=2\sqrt{\frac{\left(150\times {10}^{-8}\text{Ω}\cdot \text{m}\right)\left(3.00\text{m}\right)}{\left(3.14\right)\left(40\Omega \right)}}\\ =3.78\times {10}^{-4}\text{m}\end{array}$

Conclusion: Therefore, the diameter of wire is $3.78\times {10}^{-4}\text{m}$