November 15, 2019

Solution Set 2

Physics 208-02 – Introductory Physics II
Solution Set 2
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

§ Chapter 14, Problem 5

A water heater can generate 32,000\mathrm{kJ}/\mathrm{h}. How much water can it heat from 15^{0}\mathrm{C} to 50^{0}\mathrm{C} per hour.

In one hour the amount of heat available will be

\displaystyle Q \displaystyle= \displaystyle P\Delta t
\displaystyle= \displaystyle\left(32,000\mathrm{kJ}/\mathrm{h}\right)(1\;\mathrm{h})
\displaystyle= \displaystyle 32,000\mathrm{kJ}

We will equate this to the heat required to warm a mass m or water as

\displaystyle Q \displaystyle= \displaystyle m_{{H_{2}O}}c_{{H_{2}O}}\Delta T
\displaystyle 32,000\mathrm{kJ} \displaystyle= \displaystyle m_{{H_{2}O}}\left(4.186\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(50^{0}\mathrm{C}-15^{0}\mathrm{C}\right)
\displaystyle m_{{H_{2}O}} \displaystyle= \displaystyle 220\;\mathrm{kg}

I have expressed the answer to only two significant figures, but note that the author expressed 50^{0}\mathrm{C} to only one significant figure so actually we should round one more place.

§ Chapter 14, Problem 6

A small immersion heater is rated at 350\;\mathrm{W}. Estimate how long it will take to heat a cup of soup (assuming it is 250\;\mathrm{ml} of water) from 20^{0}\mathrm{C} to 60^{0}\mathrm{C}.

Let’s begin by calculating the required heat.

\displaystyle Q \displaystyle= \displaystyle m_{{H_{2}O}}c_{{H_{2}O}}\Delta T
\displaystyle= \displaystyle\left(250\mathrm{ml}\times\frac{1\;\mathrm{g}}{\mathrm{ml}}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{g}}\right)\left(4.186\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(60^{0}\mathrm{C}-20^{0}\mathrm{C}\right)
\displaystyle= \displaystyle 42\;\mathrm{kJ}

We now calculate the time using

\displaystyle Q \displaystyle= \displaystyle P\Delta t
\displaystyle\Delta t \displaystyle= \displaystyle Q/P
\displaystyle= \displaystyle\left(42,000\;\mathrm{J}\right)/(350\;\mathrm{W})
\displaystyle= \displaystyle 120\;\mathrm{s}

§ Chapter 14, Problem 11

A 35\;\mathrm{g} glass thermometer reads 21.6^{0}\mathrm{C} before it is placed in 135\;\mathrm{ml} of water. When the water and the thermometer come to equilibrium, the thermometer reads 39.2^{0}\mathrm{C}. What was the original temperature of the water.

Heat is flowing from the water into the glass thermometer. The sum of the two heats will total zero.

\displaystyle 0 \displaystyle= \displaystyle Q_{{glass}}+Q_{{H_{2}O}}
\displaystyle= \displaystyle m_{{glass}}c_{{glass}}\Delta T_{{glass}}+m_{{H_{2}O}}c_{{H_{2}O}}\Delta T_{{H_{2}O}}
\displaystyle= \displaystyle(0.035\;\mathrm{kg})\left(840\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(39.2^{0}\mathrm{C}-21.6^{0}\mathrm{C}\right)
\displaystyle\mbox{}+135\;\mathrm{ml}\times\frac{1\;\mathrm{g}}{\mathrm{ml}}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{g}}\left(4186\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(39.2^{0}\mathrm{C}-T\right)
\displaystyle= \displaystyle 5\underline{1}7\;\mathrm{J}+565\frac{\mathrm{J}}{\mbox{}^{0}\mathrm{C}}\left(39.2^{0}\mathrm{C}-T\right)
\displaystyle= \displaystyle 22,\underline{6}7\;\mathrm{J}-565\frac{\mathrm{J}}{\mbox{}^{0}\mathrm{C}}T
\displaystyle T \displaystyle= \displaystyle 40.1^{0}\mathrm{C}

§ Chapter 14, Problem 12

What will be the equilibrium temperature when a 245\;\mathrm{g} block of copper at 285^{0}\mathrm{C} is placed in a 145\;\mathrm{g} aluminum calorimeter cup containing 825\;\mathrm{g} of water at 12.0^{0}\mathrm{C}?

In this case we have three terms to include with heat flowing from the copper into the water and aluminum.

\displaystyle 0 \displaystyle= \displaystyle m_{{Cu}}c_{{Cu}}\Delta T_{{Cu}}+m_{{Al}}c_{{Al}}\Delta T_{{Al}}+m_{{H_{2}O}}c_{{H_{2}O}}\Delta T_{{H_{2}O}}
\displaystyle= \displaystyle(0.245\;\mathrm{kg})\left(390\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(T-285^{0}\mathrm{C}\right)
\displaystyle\mbox{}+(0.145\;\mathrm{kg})\left(900\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(T-12.0^{0}\mathrm{C}\right)
\displaystyle\mbox{}+(0.825\;\mathrm{kg})\left(4186\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(T-12.0^{0}\mathrm{C}\right)
\displaystyle= \displaystyle 3680\frac{\mathrm{J}}{\mbox{}^{0}\mathrm{C}}\; T-70,200\;\mathrm{J}
\displaystyle T \displaystyle= \displaystyle 19.1^{0}\mathrm{C}

§ Chapter 14, Problem 15

How long does it take a 750\;\mathrm{W} coffee pot to bring to a boil 0.75\;\ell of water initially at 8.0^{0}\mathrm{C}? Assume that the part of the pot that is heated with the water is made of 360\;\mathrm{g} of aluminum and that no water boils away.

First compute the heat required.

\displaystyle Q \displaystyle= \displaystyle m_{{Al}}c_{{Al}}\Delta T_{{Al}}+m_{{H_{2}O}}c_{{H_{2}O}}\Delta T_{{H_{2}O}}
\displaystyle= \displaystyle(0.360\;\mathrm{kg})\left(900\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(100.0^{0}\mathrm{C}-8.0^{0}\mathrm{C}\right)
\displaystyle\mbox{}+(0.75\;\mathrm{kg})\left(4186\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(100.0^{0}\mathrm{C}-8.0^{0}\mathrm{C}\right)
\displaystyle= \displaystyle 319,000\;\mathrm{J}

We now calculate the time using

\displaystyle Q \displaystyle= \displaystyle P\Delta t
\displaystyle\Delta t \displaystyle= \displaystyle Q/P
\displaystyle= \displaystyle\left(319,000\;\mathrm{J}\right)/(750\;\mathrm{W})
\displaystyle= \displaystyle 420\;\mathrm{s}
\displaystyle= \displaystyle 7.0\;\mathrm{minutes}

§ Chapter 14, Problem 18

The 1.20\;\mathrm{kg} head of a hammer has a speed of 6.5\;\mathrm{m}/\mathrm{s} just before it strikes a nail and is brought to rest. Estimate the temperature rise of a 14\;\mathrm{g} iron nail generated by 10 such hammer blows done in quick succession. Assume the nail absorbs all the energy.

The kinetic energy of the hammer is

\displaystyle KE \displaystyle= \displaystyle\frac{1}{2}mv^{2}
\displaystyle= \displaystyle\frac{1}{2}(1.20\;\mathrm{kg})\left(6.5\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}
\displaystyle= \displaystyle 2\underline{5}.4\;\mathrm{J}

The assumption is that this amount of kinetic energy is transfered to the nail ten (10) times in quick succession. If all of this goes into heating the nail, then

\displaystyle 10(2\underline{5}.4\;\mathrm{J}) \displaystyle= \displaystyle(0.014\;\mathrm{kg})\left(450\frac{\mathrm{J}}{\mathrm{kg}^{0}\mathrm{C}}\right)\Delta T
\displaystyle\Delta T \displaystyle= \displaystyle 40.^{0}\mathrm{C}

§ Chapter 14, Problem 21

How much heat is needed to melt 16.50\;\mathrm{kg} of silver that is initially at 20^{0}\mathrm{C}?

We need to include the heat to first elevate the temperature of the silver to its melting point of 961^{0}\mathrm{C} and add the amount of heat for the phase transition.

\displaystyle Q \displaystyle= \displaystyle m_{{Ag}}c_{{Ag}}\Delta T+m_{{Ag}}L_{{Ag}}
\displaystyle= \displaystyle(16.50\;\mathrm{kg})\left(0.23\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(961^{0}\mathrm{C}-20^{0}\mathrm{C}\right)+(16.50\;\mathrm{kg})\left(88\frac{\mathrm{kJ}}{\mathrm{kg}}\right)
\displaystyle= \displaystyle 5,\underline{0}00\;\mathrm{kJ}

§ Chapter 14, Problem 25

A cube of ice is taken from the freezer at -8.5\;\mathrm{C} and placed in a 95\;\mathrm{g} aluminum calorimeter filled with 310\;\mathrm{g} of water at room temperature of 20.0^{0}\mathrm{C}. The final situation is observed to be all water at 17.0^{0}\mathrm{C}. What was the mass of the ice cube?

This ice must first warm to its melting point, it must then go through the phase transition, and then the resulting liquid must warm to the final temperature. The original water plus the aluminum calorimeter must both cool to the final temperature. That’s a total of five terms!

\displaystyle 0 \displaystyle= \displaystyle m_{{ice}}c_{{ice}}\left(0^{0}\mathrm{C}-(-)8.5^{0}\mathrm{C}\right)+m_{{ice}}L^{\mathrm{fusion}}_{{H_{2}O}}+m_{{ice}}c_{{H_{2}O}}\left(17^{0}\mathrm{C}-0^{0}\mathrm{C}\right)
\displaystyle\mbox{}+m_{{Al}}c_{{Al}}\left(17^{0}\mathrm{C}-20^{0}\mathrm{C}\right)+m_{{H_{2}O}}c_{{H_{2}O}}\left(17^{0}\mathrm{C}-20^{0}\mathrm{C}\right)
\displaystyle= \displaystyle m_{{ice}}\left(2.100\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(0^{0}\mathrm{C}-(-)8.5^{0}\mathrm{C}\right)
\displaystyle\mbox{}+m_{{ice}}\left(333\frac{\mathrm{kJ}}{\mathrm{kg}}\right)
\displaystyle\mbox{}+m_{{ice}}\left(4.186\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(17^{0}\mathrm{C}-0^{0}\mathrm{C}\right)
\displaystyle\mbox{}+(0.095\;\mathrm{kg})\left(0.90\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(17^{0}\mathrm{C}-20^{0}\mathrm{C}\right)
\displaystyle\mbox{}+(0.31\;\mathrm{kg})\left(4.186\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(17^{0}\mathrm{C}-20^{0}\mathrm{C}\right)
\displaystyle= \displaystyle m_{{ice}}\left(422\frac{\mathrm{kJ}}{\mathrm{kg}}\right)-4.15\;\mathrm{kJ}
\displaystyle m_{{ice}} \displaystyle= \displaystyle 9.8\;\mathrm{g}

§ Chapter 14, Problem 28

What mass of steam at 100^{0}\mathrm{C} must be added to 1.00\;\mathrm{kg} of ice at 0^{0}\mathrm{C} to yield liquid water at 20^{0}\mathrm{C}?

This problem will have four terms. The steam must condense, and the resulting water must cool. The ice must melt, and the resulting water must warm.

\displaystyle 0 \displaystyle= \displaystyle-m_{{steam}}L^{{vaporization}}_{{H_{2}O}}+m_{{steam}}c_{{H_{2}O}}\left(20^{0}\mathrm{C}-100^{0}\mathrm{C}\right)
\displaystyle\mbox{}+m_{{ice}}L^{{fusion}}_{{H_{2}O}}+m_{{ice}}c_{{H_{2}O}}\left(20^{0}\mathrm{C}-0^{0}\mathrm{C}\right)
\displaystyle= \displaystyle-m_{{steam}}\left(2260\frac{\mathrm{kJ}}{\mathrm{kg}}\right)
\displaystyle\mbox{}+m_{{steam}}\left(4.186\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(20^{0}\mathrm{C}-100^{0}\mathrm{C}\right)
\displaystyle\mbox{}+(1.00\;\mathrm{kg})\left(333\frac{\mathrm{kJ}}{\mathrm{kg}}\right)
\displaystyle\mbox{}+(1.00\;\mathrm{kg})\left(4.186\frac{\mathrm{kJ}}{\mathrm{kg}^{0}\mathrm{C}}\right)\left(20^{0}\mathrm{C}-0^{0}\mathrm{C}\right)
\displaystyle= \displaystyle-m_{{steam}}\left(25\underline{9}5\frac{\mathrm{kJ}}{\mathrm{kg}}\right)+417\;\mathrm{kJ}
\displaystyle m_{{steam}} \displaystyle= \displaystyle 0.160\;\mathrm{kg}
\displaystyle= \displaystyle 160.\;\mathrm{g}

§ Chapter 14, Problem 30

A 70\;\mathrm{g} bullet traveling at 250\;\mathrm{m}/\mathrm{s} penetrates a block of ice at 0^{0}\mathrm{C} and comes to rest within the ice. Assuming that the temperature of the bullet doesn’t change appreciably, how much ice is melted as a result of the collision.

The idea is that the kinetic energy of the bullet is completely converted to heat which goes into melting ice.

\displaystyle\frac{1}{2}m_{{\mathrm{bullet}}}v^{2} \displaystyle= \displaystyle m_{\mathrm{ice}}L^{\mathrm{fusion}}_{{H_{2}O}}
\displaystyle\frac{1}{2}(0.070\;\mathrm{kg})\left(250\frac{\mathrm{m}}{\mathrm{s}}\right)^{2} \displaystyle= \displaystyle m_{\mathrm{ice}}\left(333\frac{\mathrm{kJ}}{\mathrm{kg}}\right)
\displaystyle m_{\mathrm{ice}} \displaystyle= \displaystyle 7\;\mathrm{g}

 

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