Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 21, Problem 20PQ

From Table 21.1, the specific heat of milk is 3.93 × 103 J/ (kg ∙ K). and the specific heat of water is 4.19 × 103 J/(kg ∙ K). Suppose you wish to make a large mug (0.500 L) of hot chocolate. Each liquid is initially at 5.00°C. and you need to raise their temperature to 80.0°C. The density of milk is about 1.03 × 103 kg/m3, and the density of water is 1.00 × 103 kg/m3. a. How much heat must be transferred in each case? b. If you use a small electric hot plate that puts out 455 W, how long would it take to heat each liquid?

(a)

Expert Solution
Check Mark
To determine

The amount of heat transferred to milk and water.

Answer to Problem 20PQ

The amount of heat transferred to milk is Qm=1.52×105J and water is Qw=1.57×105J.

Explanation of Solution

Write the expression for heat energy transferred to the system.

  Q=mcΔT                                                                                                     (I)

Here, Q is the heat energy, m is the mass, c is the specific heat capacity of the object, and ΔT is the change in temperature.

Write the expression for mass of the milk, relating density and volume.

  mm=ρmVm                                                                                                    (II)

Here, mm is the mass of the milk, ρm is the density of milk, and Vm is the volume of the milk container.

Write the expression for mass of the water, relating density and volume.

  mw=ρwVw                                                                                                    (III)

Here, mw is the mass of the water, ρw is the density of water, and Vw is the volume of the water container.

Rearrange the equation (I) to calculate the heat required for milk.

  Qm=mmcmΔT                                                                                             (IV)

Here, Qm is the heat energy required for milk and cm is the specific heat capacity of milk.

Rearrange the equation (I) to calculate the heat required for water.

  Qw=mwcwΔT                                                                                              (V)

Here, Qw is the heat energy required for water and cw is the specific heat capacity of water.

Conclusion:

Substitute 1.03×103kg/m3 for ρm and 0.500L for Vm in equation (II) to find ρm.

  mm=(1.03×103kg/m3)(0.500L)(0.001m31.0L)=0.515kg

Substitute 1.00×103kg/m3 for ρw and 0.500L for Vw in equation (III) to find ρw.

  mw=(1.00×103kg/m3)(0.500L)(0.001m31.0L)=0.500kg

Substitute 0.515kg for mm, 3.93×103J/kgK for cm, and 80.0°C5.00°C for ΔT in equation (IV) to find Qm.

  Qm=(0.515kg)(3.93×103J/kgK)(80.0°C5.00°C)=(0.515kg)(3.93×103J/kgK)(75.0°C)=1.52×105J

Substitute 0.500kg for mw, 4.19×103J/kgK for cw, and 80.0°C5.00°C for ΔT in equation (V) to find Qw.

  Qw=(0.500kg)(4.19×103J/kgK)(80.0°C5.00°C)=(0.500kg)(4.19×103J/kgK)(75.0°C)=1.57×105J

Therefore, the amount of heat transferred to milk is Qm=1.52×105J and water is Qw=1.57×105J.

(b)

Expert Solution
Check Mark
To determine

The time taken to heat milk and water.

Answer to Problem 20PQ

The time taken to heat milk is 3.34×102s and water is 3.45×102s.

Explanation of Solution

Write the expression for power (energy per unit time).

  P=QΔt

Here, P is the power of the electric hot plate and Δt is the time taken to heat.

Rearrange the above equation for Δt.

  Δt=QP                                                                                                                  (VI)

Rearrange the equation for time taken to heat the milk.

  Δtm=QmP                                                                                                            (VII)

Here, Δtm is the time taken to heat the milk.

Rearrange the equation for time taken to heat the water.

  Δtw=QwP                                                                                                            (VIII)

Here, Δtw is the time taken to heat the water.

Conclusion:

Substitute 1.52×105J for Qm and 455W for P in equation (VII) to find Δtm.

  Δtm=1.52×105J455W=3.34×102s

Substitute 1.57×105J for Qw and 455W for P in equation (VIII) to find Δtw.

  Δtw=1.57×105J455W=3.45×102s

Therefore, the time taken to heat milk is 3.34×102s and water is 3.45×102s.

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Chapter 21 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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