91Ó°ÊÓ

A 6-cm-diameter 13-cm-high canned drink ( \(\rho=\) \(\left.977 \mathrm{~kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(25^{\circ} \mathrm{C}\) is to be cooled to \(5^{\circ} \mathrm{C}\) by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=\) \(301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the drink to \(\operatorname{cool}\) to \(5^{\circ} \mathrm{C}\). Assume the can is agitated in water and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

Step by step solution

01

Convert values to SI units

It's always a good idea to convert the given values to SI units before proceeding with calculations. Radius of the can: \(r = \frac{6}{2} cm = 0.03 m \) Total Surface Area of the drink: \(A_s = 301.6 cm^2 = 0.03016 m^2 \) Volume of the drink: \(V = 367.6 cm^3 = 3.676 \times 10^{-4} m^3 \) Heat transfer coefficient: \(h = 120 W/(m^2 \cdot K)\)

02

Calculating the mass of the drink

We know the volume (\(V\)) and the density (\(\rho\)) of the drink, to find the mass, use the equation: $$ m = \rho V $$ $$ m = 0.977 \times 3.676 \times 10^{-3} $$ $$ m = 3.694 kg $$

03

Calculate the heat transfer required to reach 5°C

The heat transfer, \(Q\), required to change the temperature of the drink from \(25°C\) to \(5°C\) can be calculated using the formula: $$ Q = m \times c_p \times \Delta T $$ Here, \(c_p\) is the specific heat capacity and \(\Delta T\) is the change in temperature. $$ Q = 3.694 \times 4180 \times (25 - 5) $$ $$ Q = 3.694 \times 4180 \times 20 $$ $$ Q = 30,7412 J $$

04

Apply Newton's Law of Cooling

Newton's Law of Cooling states: $$ Q = hA_s \int_{0}^{t} (T - T_{s}) dt $$ Since the drink cools uniformly, \(T - T_s\) remains constant and the integral can be evaluated: $$ Q = h A_s (T - T_s) t $$

05

Solve for time, t

Rearrange the equation to find the time it takes for the drink to cool down to 5°C. $$ t = \frac{Q}{h A_s (T - T_s)} $$ $$ t = \frac{307412}{120 \times 0.03016 \times (5 - 0)} $$ $$ t = 525.84s $$ Now, convert seconds to minutes: $$ t = 525.84s \times \frac{1 \text{ min}}{60 s} \approx 8.76 \text{ min} $$ Therefore, the answer is (b) \(8.7 \text{ min}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer Coefficient

When a refreshing canned drink is plunged into a chilly bath of iced water, it's the heat transfer coefficient that plays a pivotal role in determining the swiftness of cooling. But what exactly is the heat transfer coefficient, you might wonder? In simple terms, it represents how effectively heat can be transported from the can to the water that surrounds it.

The coefficient's value depends on various factors, including the nature of the convection process (which could be natural or forced) and the properties of the fluids in contact. A higher heat transfer coefficient implies a more efficient transfer of heat from the drink to the iced water. So when we talk about the 'h' in the textbook problem, think of it as a numerical indicator of just how good the water is at whisking heat away from your summertime beverage.

Specific Heat Capacity

Have you ever wondered why some things get hot real quick and others don't? This is where the concept of specific heat capacity comes into play. It is a measure of how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius.

For your canned drink, the specific heat capacity tells us just how much warmth the liquid holds onto before it's ready to let go and cool down. In the step-by-step solution, this concept is crucial in calculating the total amount of thermal energy the drink initially possesses and how much it needs to lose to reach the desired cooler temperature. Different substances have different specific heat capacities, so while your drink is cooling off, remember that it's the drink's specific heat capacity that dictates how long you'll have to wait before taking that refreshing sip.

Thermal Energy Conservation

Conserving thermal energy sounds a bit like financial advising, doesn't it? But rather than saving money, it's about understanding how heat energy is preserved and distributed in physical systems. In our textbook example, the principle of thermal energy conservation ensures that all the heat leaving the can is being transferred into the surrounding water.

By applying this principle, we can trace the journey of every joule of energy as the drink cools down to the targeted temperature. It allows us to set up an equation equating the loss of heat from the drink to the heat gained by the water (even though in this case, the icy bath doesn't get much warmer). This conservation principle acts like a meticulous accountant, making sure that the heat energy balance sheet tallies up perfectly.